Mixing
Maximizing $ (4a-3b)^2+(5b-4c)^2+(3c-5a)^2$, such that $a^2+b^2+c^2=1 $
$begingroup$
If $$a^2+b^2+c^2=1 $$here a,b,c are the real numbers then find the maximum value of $$ (4a-3b)^2+(5b-4c)^2+(3c-5a)^2$$
I tried to think with vectors, that is direction cosines of lines.
But then the expression is not getting simplified.
inequality optimization quadratics maxima-minima
edited Jan 27 at 6:38
Michael Rozenberg
109k1895200
asked Jan 27 at 2:41
Math_centricMath_centric
233
$endgroup$
add a comment |
$begingroup$
If $$a^2+b^2+c^2=1 $$here a,b,c are the real numbers then find the maximum value of $$ (4a-3b)^2+(5b-4c)^2+(3c-5a)^2$$
I tried to think with vectors, that is direction cosines of lines.
But then the expression is not getting simplified.
inequality optimization quadratics maxima-minima
edited Jan 27 at 6:38
Michael Rozenberg
109k1895200
asked Jan 27 at 2:41
Math_centricMath_centric
233
$endgroup$
1
$begingroup$
Hint: the second quadratic form is clearly positive definite and associated to a symmetric matrix, whose real eigenvalues describe...
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:49
$begingroup$
You’re looking for a norm of an obvious matrix...
$endgroup$
– Macavity
Jan 27 at 5:52
add a comment |
$begingroup$
If $$a^2+b^2+c^2=1 $$here a,b,c are the real numbers then find the maximum value of $$ (4a-3b)^2+(5b-4c)^2+(3c-5a)^2$$
I tried to think with vectors, that is direction cosines of lines.
But then the expression is not getting simplified.
inequality optimization quadratics maxima-minima
edited Jan 27 at 6:38
Michael Rozenberg
109k1895200
asked Jan 27 at 2:41
Math_centricMath_centric
233
$endgroup$
If $$a^2+b^2+c^2=1 $$here a,b,c are the real numbers then find the maximum value of $$ (4a-3b)^2+(5b-4c)^2+(3c-5a)^2$$
I tried to think with vectors, that is direction cosines of lines.
But then the expression is not getting simplified.
inequality optimization quadratics maxima-minima
inequality optimization quadratics maxima-minima
edited Jan 27 at 6:38
Michael Rozenberg
109k1895200
asked Jan 27 at 2:41
Math_centricMath_centric
233
edited Jan 27 at 6:38
Michael Rozenberg
109k1895200
asked Jan 27 at 2:41
Math_centricMath_centric
233
edited Jan 27 at 6:38
Michael Rozenberg
109k1895200
edited Jan 27 at 6:38
Michael Rozenberg
109k1895200
edited Jan 27 at 6:38
Michael Rozenberg
109k1895200
109k1895200
asked Jan 27 at 2:41
Math_centricMath_centric
233
asked Jan 27 at 2:41
Math_centricMath_centric
233
asked Jan 27 at 2:41
Math_centricMath_centric
233
233
1
$begingroup$
Hint: the second quadratic form is clearly positive definite and associated to a symmetric matrix, whose real eigenvalues describe...
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:49
$begingroup$
You’re looking for a norm of an obvious matrix...
$endgroup$
– Macavity
Jan 27 at 5:52
add a comment |
1
$begingroup$
Hint: the second quadratic form is clearly positive definite and associated to a symmetric matrix, whose real eigenvalues describe...
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:49
$begingroup$
You’re looking for a norm of an obvious matrix...
$endgroup$
– Macavity
Jan 27 at 5:52
1
1
$begingroup$
Hint: the second quadratic form is clearly positive definite and associated to a symmetric matrix, whose real eigenvalues describe...
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:49
$begingroup$
Hint: the second quadratic form is clearly positive definite and associated to a symmetric matrix, whose real eigenvalues describe...
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:49
$begingroup$
You’re looking for a norm of an obvious matrix...
$endgroup$
– Macavity
Jan 27 at 5:52
$begingroup$
You’re looking for a norm of an obvious matrix...
$endgroup$
– Macavity
Jan 27 at 5:52
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
try opening the brackets.
square terms with coeeficents combine them
replace $$ a^2+b^2$$ type terms with $$c^2$$.
you have
$$ 50-(3a+4b+5c)^2$$
answered Jan 27 at 4:10
mavericmaveric
88512
$endgroup$
1
$begingroup$
this is the shortest and perfect solution.
$endgroup$
– Math_centric
Jan 29 at 11:22
add a comment |
$begingroup$
Using the method of Lagrange multipliers, consider
$$F= (4a-3b)^2+(5b-4c)^2+(3c-5a)^2+lambda(a^2+b^2+c^2-1) tag 1$$ Computing the partial derivatives
$$frac{partial F}{partial a}=8 (4 a-3 b)-10 (3 c-5 a)+2 a lambda tag 2$$
$$frac{partial F}{partial b}=-6 (4 a-3 b)+10 (5 b-4 c)+2 b lambdatag 3$$
$$frac{partial F}{partial c}=6 (3 c-5 a)-8 (5 b-4 c)+2 c lambdatag 4$$
$$frac{partial F}{partial lambda}=a^2+b^2+c^2-1tag 5$$
From $(2)$, $b=frac{1}{12} (a lambda +41 a-15 c)$; plug in $(3)$ to get $c=frac{1}{15} (a lambda +25 a)$ which makes $b=frac{4 }{3}a$; plug $b$ and $c$ in $(4)$ to get
$$frac{2}{15} a lambda (lambda +50)=0 tag 6$$ So, we have three cases : $a=0$ , $lambda=0$, $lambda=-50$.
The first case $a=0$ can be discarded since it would make $a=b=c=0$ which does not satisfy the constraint.
The second case $lambda=0$ would make $c=frac{5 }{3}a$ which, in turn, would make $frac{50 }{9}a^2=1$ from the constraint and then the value of the expression to maximize would just be $0$.
So, what is left is the case $lambda=-50$ which makes $c=-frac{5 }{3}a$ which gives again $frac{50 }{9}a^2=1$ from the constraint. The expression to maximize if then $frac{a^2 lambda ^2}{9}$ with $lambda=-50$ and $frac{50 }{9}a^2=1$ which then gives a maximum vzlue of $50$.
I let you finishing what could be required.
answered Jan 27 at 4:13
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
add a comment |
$begingroup$
We need to find a minimal value of $k$ for which the following inequality is true for all reals $a$, $b$ and $c$.
$$(4a-3b)^2+(5b-4c)^2+(3c-5a)^2leq k(a^2+b^2+c^2)$$ or
$$(k-41)a^2+(k-34)b^2+(k-25)c^2+24ab+40bc+30acgeq0$$ or
$$(k-41)a^2+6(4b+5c)a+(k-34)b^2+40bc+(k-25)c^2geq0,$$ for which we need
$$k>41$$ and
$$9(4b+5c)^2-(k-41)((k-34)b^2+40bc+(k-25)c^2)leq0$$ or
$$(k^2-75k+1250)b^2+(k^2-66k+800)c^2+40(k-50)bcgeq0$$ or
$$(k-50)((k-25)b^2+(k-16)c^2+40bc)geq0.$$
But $$(k-25)b^2+(k-16)c^2+40bcgeq16b^2+25c^2+40bc=(4b+5c)^2geq0,$$ which gives $$kgeq50.$$
For $k=50$ we obtain $$9a^2+6(4b+5c)a+(4b+5c)^2geq0$$ or
$$(3a+4b+5c)^2geq0,$$ which gives that the equality occurs for $$3a+4b+5c=0$$ and $$a^2+b^2+c^2=1,$$
which says that $50$ is a maximal value.
answered Jan 27 at 6:37
Michael RozenbergMichael Rozenberg
109k1895200
$endgroup$
add a comment |
$begingroup$
Use the technique of Lagrange multipliers to solve.
Essentially, if $$f(a,b,c) = a^2 + b^2 + c^2,$$ $$g(a,b,c) = (4a-3b)^2+(5b-4c)^2+(3c-5a)^2,$$ then find the values $(a,b,c)$ such that $vec nabla f$ is parallel to $vec nabla g$.
Then, by a bit of testing on your solutions, select the one that satisfies $f(a,b,c) = 1$ and is a maximum of $g(a,b,c)$ (as opposed to a minimum or saddle point, for instance).
answered Jan 27 at 3:01
Trevor KafkaTrevor Kafka
5309
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add a comment |
$begingroup$
Assume $vec{A}=ahat{i}+bhat{j}+chat{k}$ and $vec{B}=3hat{i}+4hat{j}+5hat{k}$
$$bigg|vec{A}times vec{B}bigg|^2=bigg|vec{A}bigg|^2bigg|vec{B}bigg|^2-bigg(vec{A}cdotvec{B}bigg)^2$$
$$(4a-3b)^2+(5b-4c)^2+(3c-5a)^2=50-(3a+4b+5c)^2leq 50$$
answered Feb 4 at 7:10
jackyjacky
1,259816
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
try opening the brackets.
square terms with coeeficents combine them
replace $$ a^2+b^2$$ type terms with $$c^2$$.
you have
$$ 50-(3a+4b+5c)^2$$
answered Jan 27 at 4:10
mavericmaveric
88512
$endgroup$
1
$begingroup$
this is the shortest and perfect solution.
$endgroup$
– Math_centric
Jan 29 at 11:22
add a comment |
$begingroup$
try opening the brackets.
square terms with coeeficents combine them
replace $$ a^2+b^2$$ type terms with $$c^2$$.
you have
$$ 50-(3a+4b+5c)^2$$
answered Jan 27 at 4:10
mavericmaveric
88512
$endgroup$
1
$begingroup$
this is the shortest and perfect solution.
$endgroup$
– Math_centric
Jan 29 at 11:22
add a comment |
$begingroup$
try opening the brackets.
square terms with coeeficents combine them
replace $$ a^2+b^2$$ type terms with $$c^2$$.
you have
$$ 50-(3a+4b+5c)^2$$
answered Jan 27 at 4:10
mavericmaveric
88512
$endgroup$
try opening the brackets.
square terms with coeeficents combine them
replace $$ a^2+b^2$$ type terms with $$c^2$$.
you have
$$ 50-(3a+4b+5c)^2$$
answered Jan 27 at 4:10
mavericmaveric
88512
answered Jan 27 at 4:10
mavericmaveric
88512
answered Jan 27 at 4:10
mavericmaveric
88512
answered Jan 27 at 4:10
mavericmaveric
88512
88512
1
$begingroup$
this is the shortest and perfect solution.
$endgroup$
– Math_centric
Jan 29 at 11:22
add a comment |
1
$begingroup$
this is the shortest and perfect solution.
$endgroup$
– Math_centric
Jan 29 at 11:22
1
1
$begingroup$
this is the shortest and perfect solution.
$endgroup$
– Math_centric
Jan 29 at 11:22
$begingroup$
this is the shortest and perfect solution.
$endgroup$
– Math_centric
Jan 29 at 11:22
add a comment |
$begingroup$
Using the method of Lagrange multipliers, consider
$$F= (4a-3b)^2+(5b-4c)^2+(3c-5a)^2+lambda(a^2+b^2+c^2-1) tag 1$$ Computing the partial derivatives
$$frac{partial F}{partial a}=8 (4 a-3 b)-10 (3 c-5 a)+2 a lambda tag 2$$
$$frac{partial F}{partial b}=-6 (4 a-3 b)+10 (5 b-4 c)+2 b lambdatag 3$$
$$frac{partial F}{partial c}=6 (3 c-5 a)-8 (5 b-4 c)+2 c lambdatag 4$$
$$frac{partial F}{partial lambda}=a^2+b^2+c^2-1tag 5$$
From $(2)$, $b=frac{1}{12} (a lambda +41 a-15 c)$; plug in $(3)$ to get $c=frac{1}{15} (a lambda +25 a)$ which makes $b=frac{4 }{3}a$; plug $b$ and $c$ in $(4)$ to get
$$frac{2}{15} a lambda (lambda +50)=0 tag 6$$ So, we have three cases : $a=0$ , $lambda=0$, $lambda=-50$.
The first case $a=0$ can be discarded since it would make $a=b=c=0$ which does not satisfy the constraint.
The second case $lambda=0$ would make $c=frac{5 }{3}a$ which, in turn, would make $frac{50 }{9}a^2=1$ from the constraint and then the value of the expression to maximize would just be $0$.
So, what is left is the case $lambda=-50$ which makes $c=-frac{5 }{3}a$ which gives again $frac{50 }{9}a^2=1$ from the constraint. The expression to maximize if then $frac{a^2 lambda ^2}{9}$ with $lambda=-50$ and $frac{50 }{9}a^2=1$ which then gives a maximum vzlue of $50$.
I let you finishing what could be required.
answered Jan 27 at 4:13
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
add a comment |
$begingroup$
Using the method of Lagrange multipliers, consider
$$F= (4a-3b)^2+(5b-4c)^2+(3c-5a)^2+lambda(a^2+b^2+c^2-1) tag 1$$ Computing the partial derivatives
$$frac{partial F}{partial a}=8 (4 a-3 b)-10 (3 c-5 a)+2 a lambda tag 2$$
$$frac{partial F}{partial b}=-6 (4 a-3 b)+10 (5 b-4 c)+2 b lambdatag 3$$
$$frac{partial F}{partial c}=6 (3 c-5 a)-8 (5 b-4 c)+2 c lambdatag 4$$
$$frac{partial F}{partial lambda}=a^2+b^2+c^2-1tag 5$$
From $(2)$, $b=frac{1}{12} (a lambda +41 a-15 c)$; plug in $(3)$ to get $c=frac{1}{15} (a lambda +25 a)$ which makes $b=frac{4 }{3}a$; plug $b$ and $c$ in $(4)$ to get
$$frac{2}{15} a lambda (lambda +50)=0 tag 6$$ So, we have three cases : $a=0$ , $lambda=0$, $lambda=-50$.
The first case $a=0$ can be discarded since it would make $a=b=c=0$ which does not satisfy the constraint.
The second case $lambda=0$ would make $c=frac{5 }{3}a$ which, in turn, would make $frac{50 }{9}a^2=1$ from the constraint and then the value of the expression to maximize would just be $0$.
So, what is left is the case $lambda=-50$ which makes $c=-frac{5 }{3}a$ which gives again $frac{50 }{9}a^2=1$ from the constraint. The expression to maximize if then $frac{a^2 lambda ^2}{9}$ with $lambda=-50$ and $frac{50 }{9}a^2=1$ which then gives a maximum vzlue of $50$.
I let you finishing what could be required.
answered Jan 27 at 4:13
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
add a comment |
$begingroup$
Using the method of Lagrange multipliers, consider
$$F= (4a-3b)^2+(5b-4c)^2+(3c-5a)^2+lambda(a^2+b^2+c^2-1) tag 1$$ Computing the partial derivatives
$$frac{partial F}{partial a}=8 (4 a-3 b)-10 (3 c-5 a)+2 a lambda tag 2$$
$$frac{partial F}{partial b}=-6 (4 a-3 b)+10 (5 b-4 c)+2 b lambdatag 3$$
$$frac{partial F}{partial c}=6 (3 c-5 a)-8 (5 b-4 c)+2 c lambdatag 4$$
$$frac{partial F}{partial lambda}=a^2+b^2+c^2-1tag 5$$
From $(2)$, $b=frac{1}{12} (a lambda +41 a-15 c)$; plug in $(3)$ to get $c=frac{1}{15} (a lambda +25 a)$ which makes $b=frac{4 }{3}a$; plug $b$ and $c$ in $(4)$ to get
$$frac{2}{15} a lambda (lambda +50)=0 tag 6$$ So, we have three cases : $a=0$ , $lambda=0$, $lambda=-50$.
The first case $a=0$ can be discarded since it would make $a=b=c=0$ which does not satisfy the constraint.
The second case $lambda=0$ would make $c=frac{5 }{3}a$ which, in turn, would make $frac{50 }{9}a^2=1$ from the constraint and then the value of the expression to maximize would just be $0$.
So, what is left is the case $lambda=-50$ which makes $c=-frac{5 }{3}a$ which gives again $frac{50 }{9}a^2=1$ from the constraint. The expression to maximize if then $frac{a^2 lambda ^2}{9}$ with $lambda=-50$ and $frac{50 }{9}a^2=1$ which then gives a maximum vzlue of $50$.
I let you finishing what could be required.
answered Jan 27 at 4:13
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
Using the method of Lagrange multipliers, consider
$$F= (4a-3b)^2+(5b-4c)^2+(3c-5a)^2+lambda(a^2+b^2+c^2-1) tag 1$$ Computing the partial derivatives
$$frac{partial F}{partial a}=8 (4 a-3 b)-10 (3 c-5 a)+2 a lambda tag 2$$
$$frac{partial F}{partial b}=-6 (4 a-3 b)+10 (5 b-4 c)+2 b lambdatag 3$$
$$frac{partial F}{partial c}=6 (3 c-5 a)-8 (5 b-4 c)+2 c lambdatag 4$$
$$frac{partial F}{partial lambda}=a^2+b^2+c^2-1tag 5$$
From $(2)$, $b=frac{1}{12} (a lambda +41 a-15 c)$; plug in $(3)$ to get $c=frac{1}{15} (a lambda +25 a)$ which makes $b=frac{4 }{3}a$; plug $b$ and $c$ in $(4)$ to get
$$frac{2}{15} a lambda (lambda +50)=0 tag 6$$ So, we have three cases : $a=0$ , $lambda=0$, $lambda=-50$.
The first case $a=0$ can be discarded since it would make $a=b=c=0$ which does not satisfy the constraint.
The second case $lambda=0$ would make $c=frac{5 }{3}a$ which, in turn, would make $frac{50 }{9}a^2=1$ from the constraint and then the value of the expression to maximize would just be $0$.
So, what is left is the case $lambda=-50$ which makes $c=-frac{5 }{3}a$ which gives again $frac{50 }{9}a^2=1$ from the constraint. The expression to maximize if then $frac{a^2 lambda ^2}{9}$ with $lambda=-50$ and $frac{50 }{9}a^2=1$ which then gives a maximum vzlue of $50$.
I let you finishing what could be required.
answered Jan 27 at 4:13
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 27 at 4:13
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 27 at 4:13
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 27 at 4:13
Claude LeiboviciClaude Leibovici
124k1158135
124k1158135
add a comment |
add a comment |
$begingroup$
We need to find a minimal value of $k$ for which the following inequality is true for all reals $a$, $b$ and $c$.
$$(4a-3b)^2+(5b-4c)^2+(3c-5a)^2leq k(a^2+b^2+c^2)$$ or
$$(k-41)a^2+(k-34)b^2+(k-25)c^2+24ab+40bc+30acgeq0$$ or
$$(k-41)a^2+6(4b+5c)a+(k-34)b^2+40bc+(k-25)c^2geq0,$$ for which we need
$$k>41$$ and
$$9(4b+5c)^2-(k-41)((k-34)b^2+40bc+(k-25)c^2)leq0$$ or
$$(k^2-75k+1250)b^2+(k^2-66k+800)c^2+40(k-50)bcgeq0$$ or
$$(k-50)((k-25)b^2+(k-16)c^2+40bc)geq0.$$
But $$(k-25)b^2+(k-16)c^2+40bcgeq16b^2+25c^2+40bc=(4b+5c)^2geq0,$$ which gives $$kgeq50.$$
For $k=50$ we obtain $$9a^2+6(4b+5c)a+(4b+5c)^2geq0$$ or
$$(3a+4b+5c)^2geq0,$$ which gives that the equality occurs for $$3a+4b+5c=0$$ and $$a^2+b^2+c^2=1,$$
which says that $50$ is a maximal value.
answered Jan 27 at 6:37
Michael RozenbergMichael Rozenberg
109k1895200
$endgroup$
add a comment |
$begingroup$
We need to find a minimal value of $k$ for which the following inequality is true for all reals $a$, $b$ and $c$.
$$(4a-3b)^2+(5b-4c)^2+(3c-5a)^2leq k(a^2+b^2+c^2)$$ or
$$(k-41)a^2+(k-34)b^2+(k-25)c^2+24ab+40bc+30acgeq0$$ or
$$(k-41)a^2+6(4b+5c)a+(k-34)b^2+40bc+(k-25)c^2geq0,$$ for which we need
$$k>41$$ and
$$9(4b+5c)^2-(k-41)((k-34)b^2+40bc+(k-25)c^2)leq0$$ or
$$(k^2-75k+1250)b^2+(k^2-66k+800)c^2+40(k-50)bcgeq0$$ or
$$(k-50)((k-25)b^2+(k-16)c^2+40bc)geq0.$$
But $$(k-25)b^2+(k-16)c^2+40bcgeq16b^2+25c^2+40bc=(4b+5c)^2geq0,$$ which gives $$kgeq50.$$
For $k=50$ we obtain $$9a^2+6(4b+5c)a+(4b+5c)^2geq0$$ or
$$(3a+4b+5c)^2geq0,$$ which gives that the equality occurs for $$3a+4b+5c=0$$ and $$a^2+b^2+c^2=1,$$
which says that $50$ is a maximal value.
answered Jan 27 at 6:37
Michael RozenbergMichael Rozenberg
109k1895200
$endgroup$
add a comment |
$begingroup$
We need to find a minimal value of $k$ for which the following inequality is true for all reals $a$, $b$ and $c$.
$$(4a-3b)^2+(5b-4c)^2+(3c-5a)^2leq k(a^2+b^2+c^2)$$ or
$$(k-41)a^2+(k-34)b^2+(k-25)c^2+24ab+40bc+30acgeq0$$ or
$$(k-41)a^2+6(4b+5c)a+(k-34)b^2+40bc+(k-25)c^2geq0,$$ for which we need
$$k>41$$ and
$$9(4b+5c)^2-(k-41)((k-34)b^2+40bc+(k-25)c^2)leq0$$ or
$$(k^2-75k+1250)b^2+(k^2-66k+800)c^2+40(k-50)bcgeq0$$ or
$$(k-50)((k-25)b^2+(k-16)c^2+40bc)geq0.$$
But $$(k-25)b^2+(k-16)c^2+40bcgeq16b^2+25c^2+40bc=(4b+5c)^2geq0,$$ which gives $$kgeq50.$$
For $k=50$ we obtain $$9a^2+6(4b+5c)a+(4b+5c)^2geq0$$ or
$$(3a+4b+5c)^2geq0,$$ which gives that the equality occurs for $$3a+4b+5c=0$$ and $$a^2+b^2+c^2=1,$$
which says that $50$ is a maximal value.
answered Jan 27 at 6:37
Michael RozenbergMichael Rozenberg
109k1895200
$endgroup$
We need to find a minimal value of $k$ for which the following inequality is true for all reals $a$, $b$ and $c$.
$$(4a-3b)^2+(5b-4c)^2+(3c-5a)^2leq k(a^2+b^2+c^2)$$ or
$$(k-41)a^2+(k-34)b^2+(k-25)c^2+24ab+40bc+30acgeq0$$ or
$$(k-41)a^2+6(4b+5c)a+(k-34)b^2+40bc+(k-25)c^2geq0,$$ for which we need
$$k>41$$ and
$$9(4b+5c)^2-(k-41)((k-34)b^2+40bc+(k-25)c^2)leq0$$ or
$$(k^2-75k+1250)b^2+(k^2-66k+800)c^2+40(k-50)bcgeq0$$ or
$$(k-50)((k-25)b^2+(k-16)c^2+40bc)geq0.$$
But $$(k-25)b^2+(k-16)c^2+40bcgeq16b^2+25c^2+40bc=(4b+5c)^2geq0,$$ which gives $$kgeq50.$$
For $k=50$ we obtain $$9a^2+6(4b+5c)a+(4b+5c)^2geq0$$ or
$$(3a+4b+5c)^2geq0,$$ which gives that the equality occurs for $$3a+4b+5c=0$$ and $$a^2+b^2+c^2=1,$$
which says that $50$ is a maximal value.
answered Jan 27 at 6:37
Michael RozenbergMichael Rozenberg
109k1895200
answered Jan 27 at 6:37
Michael RozenbergMichael Rozenberg
109k1895200
answered Jan 27 at 6:37
Michael RozenbergMichael Rozenberg
109k1895200
answered Jan 27 at 6:37
Michael RozenbergMichael Rozenberg
109k1895200
109k1895200
add a comment |
add a comment |
$begingroup$
Use the technique of Lagrange multipliers to solve.
Essentially, if $$f(a,b,c) = a^2 + b^2 + c^2,$$ $$g(a,b,c) = (4a-3b)^2+(5b-4c)^2+(3c-5a)^2,$$ then find the values $(a,b,c)$ such that $vec nabla f$ is parallel to $vec nabla g$.
Then, by a bit of testing on your solutions, select the one that satisfies $f(a,b,c) = 1$ and is a maximum of $g(a,b,c)$ (as opposed to a minimum or saddle point, for instance).
answered Jan 27 at 3:01
Trevor KafkaTrevor Kafka
5309
$endgroup$
add a comment |
$begingroup$
Use the technique of Lagrange multipliers to solve.
Essentially, if $$f(a,b,c) = a^2 + b^2 + c^2,$$ $$g(a,b,c) = (4a-3b)^2+(5b-4c)^2+(3c-5a)^2,$$ then find the values $(a,b,c)$ such that $vec nabla f$ is parallel to $vec nabla g$.
Then, by a bit of testing on your solutions, select the one that satisfies $f(a,b,c) = 1$ and is a maximum of $g(a,b,c)$ (as opposed to a minimum or saddle point, for instance).
answered Jan 27 at 3:01
Trevor KafkaTrevor Kafka
5309
$endgroup$
add a comment |
$begingroup$
Use the technique of Lagrange multipliers to solve.
Essentially, if $$f(a,b,c) = a^2 + b^2 + c^2,$$ $$g(a,b,c) = (4a-3b)^2+(5b-4c)^2+(3c-5a)^2,$$ then find the values $(a,b,c)$ such that $vec nabla f$ is parallel to $vec nabla g$.
Then, by a bit of testing on your solutions, select the one that satisfies $f(a,b,c) = 1$ and is a maximum of $g(a,b,c)$ (as opposed to a minimum or saddle point, for instance).
answered Jan 27 at 3:01
Trevor KafkaTrevor Kafka
5309
$endgroup$
Use the technique of Lagrange multipliers to solve.
Essentially, if $$f(a,b,c) = a^2 + b^2 + c^2,$$ $$g(a,b,c) = (4a-3b)^2+(5b-4c)^2+(3c-5a)^2,$$ then find the values $(a,b,c)$ such that $vec nabla f$ is parallel to $vec nabla g$.
Then, by a bit of testing on your solutions, select the one that satisfies $f(a,b,c) = 1$ and is a maximum of $g(a,b,c)$ (as opposed to a minimum or saddle point, for instance).
answered Jan 27 at 3:01
Trevor KafkaTrevor Kafka
5309
answered Jan 27 at 3:01
Trevor KafkaTrevor Kafka
5309
answered Jan 27 at 3:01
Trevor KafkaTrevor Kafka
5309
answered Jan 27 at 3:01
Trevor KafkaTrevor Kafka
5309
5309
add a comment |
add a comment |
$begingroup$
Assume $vec{A}=ahat{i}+bhat{j}+chat{k}$ and $vec{B}=3hat{i}+4hat{j}+5hat{k}$
$$bigg|vec{A}times vec{B}bigg|^2=bigg|vec{A}bigg|^2bigg|vec{B}bigg|^2-bigg(vec{A}cdotvec{B}bigg)^2$$
$$(4a-3b)^2+(5b-4c)^2+(3c-5a)^2=50-(3a+4b+5c)^2leq 50$$
answered Feb 4 at 7:10
jackyjacky
1,259816
$endgroup$
add a comment |
$begingroup$
Assume $vec{A}=ahat{i}+bhat{j}+chat{k}$ and $vec{B}=3hat{i}+4hat{j}+5hat{k}$
$$bigg|vec{A}times vec{B}bigg|^2=bigg|vec{A}bigg|^2bigg|vec{B}bigg|^2-bigg(vec{A}cdotvec{B}bigg)^2$$
$$(4a-3b)^2+(5b-4c)^2+(3c-5a)^2=50-(3a+4b+5c)^2leq 50$$
answered Feb 4 at 7:10
jackyjacky
1,259816
$endgroup$
add a comment |
$begingroup$
Assume $vec{A}=ahat{i}+bhat{j}+chat{k}$ and $vec{B}=3hat{i}+4hat{j}+5hat{k}$
$$bigg|vec{A}times vec{B}bigg|^2=bigg|vec{A}bigg|^2bigg|vec{B}bigg|^2-bigg(vec{A}cdotvec{B}bigg)^2$$
$$(4a-3b)^2+(5b-4c)^2+(3c-5a)^2=50-(3a+4b+5c)^2leq 50$$
answered Feb 4 at 7:10
jackyjacky
1,259816
$endgroup$
Assume $vec{A}=ahat{i}+bhat{j}+chat{k}$ and $vec{B}=3hat{i}+4hat{j}+5hat{k}$
$$bigg|vec{A}times vec{B}bigg|^2=bigg|vec{A}bigg|^2bigg|vec{B}bigg|^2-bigg(vec{A}cdotvec{B}bigg)^2$$
$$(4a-3b)^2+(5b-4c)^2+(3c-5a)^2=50-(3a+4b+5c)^2leq 50$$
answered Feb 4 at 7:10
jackyjacky
1,259816
answered Feb 4 at 7:10
jackyjacky
1,259816
answered Feb 4 at 7:10
jackyjacky
1,259816
answered Feb 4 at 7:10
jackyjacky
1,259816
1,259816
add a comment |
add a comment |
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Hint: the second quadratic form is clearly positive definite and associated to a symmetric matrix, whose real eigenvalues describe...
$endgroup$
– Jack D'Aurizio
Jan 27 at 2:49
$begingroup$
You’re looking for a norm of an obvious matrix...
$endgroup$
– Macavity
Jan 27 at 5:52