Mixing
Suppose $x, y, z$ are positive real number such that $x + 2y + 3z = 1$. Find the maximum value of $xyz^2$
$begingroup$
I'm trying to solve this problem using Lagrange's multiplier method but I'm unable to get the value of lambda and also $z.$ kindly help me out with this problem.
I'm getting $x = 2y$
I'm stuck somewhere between the steps
maxima-minima
edited May 1 '18 at 6:52
User
1,88221229
asked May 1 '18 at 5:44
Priya WadhwaPriya Wadhwa
789
$endgroup$
add a comment |
$begingroup$
I'm trying to solve this problem using Lagrange's multiplier method but I'm unable to get the value of lambda and also $z.$ kindly help me out with this problem.
I'm getting $x = 2y$
I'm stuck somewhere between the steps
maxima-minima
edited May 1 '18 at 6:52
User
1,88221229
asked May 1 '18 at 5:44
Priya WadhwaPriya Wadhwa
789
$endgroup$
add a comment |
$begingroup$
I'm trying to solve this problem using Lagrange's multiplier method but I'm unable to get the value of lambda and also $z.$ kindly help me out with this problem.
I'm getting $x = 2y$
I'm stuck somewhere between the steps
maxima-minima
edited May 1 '18 at 6:52
User
1,88221229
asked May 1 '18 at 5:44
Priya WadhwaPriya Wadhwa
789
$endgroup$
I'm trying to solve this problem using Lagrange's multiplier method but I'm unable to get the value of lambda and also $z.$ kindly help me out with this problem.
I'm getting $x = 2y$
I'm stuck somewhere between the steps
maxima-minima
maxima-minima
edited May 1 '18 at 6:52
User
1,88221229
asked May 1 '18 at 5:44
Priya WadhwaPriya Wadhwa
789
edited May 1 '18 at 6:52
User
1,88221229
asked May 1 '18 at 5:44
Priya WadhwaPriya Wadhwa
789
edited May 1 '18 at 6:52
User
1,88221229
edited May 1 '18 at 6:52
User
1,88221229
edited May 1 '18 at 6:52
User
1,88221229
1,88221229
asked May 1 '18 at 5:44
Priya WadhwaPriya Wadhwa
789
asked May 1 '18 at 5:44
Priya WadhwaPriya Wadhwa
789
asked May 1 '18 at 5:44
Priya WadhwaPriya Wadhwa
789
789
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
By definition, $lambda$ is a constant, and we don't even really care what $lambda$ is. It is always linear and factored from the gradient, so instead of finding $lambda$, use it to jump from one equation to another.
$lambda=yz^2$
$lambda=frac{xz^2}{2}$
$lambda=frac{2xyz}{3}$
answered May 1 '18 at 6:14
AEngineerAEngineer
1,6521318
$endgroup$
1
$begingroup$
by equating values of lambda and the constraint x + 2y + 3z = 1 the values of x,y,z can be calculated and i got the answer x = 1/4, y= 1/8 and z = 1/6.
$endgroup$
– Priya Wadhwa
May 1 '18 at 6:41
add a comment |
$begingroup$
Alt. hint (no calculus): the following inequality holds by AM-GM, with equality iff $displaystyle x = 2y = frac{3}{2}z,$:
$$
frac{sqrt{3}}{sqrt[4]{2}} cdot sqrt[4]{xyz^2} = sqrt[4]{x cdot 2y cdot frac{3}{2}z cdot frac{3}{2}z} ;le; frac{x+2y+3z}{4} = frac{1}{4} ;;iff;;xyz^2 le frac{2}{3^2 cdot 4^4} = frac{1}{1152}
$$
answered May 1 '18 at 7:05
dxivdxiv
58k648102
$endgroup$
add a comment |
$begingroup$
Hint: prove that $$xyz^2le frac{1}{1152}$$ and the equal sign holds if $$x=frac{1}{4},y=frac{1}{8},z=frac{1}{6}$$
answered May 1 '18 at 5:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
$endgroup$
add a comment |
$begingroup$
You can also convert $xyz^2$ into a function of two variables $f(y,z)=(1-2y-3z)yz^2$ and use this test to conclude that the maxima occurs at $(frac{1}{4},frac{1}{8}, frac{1}{6})$ and is $1152.$ You can ignore the other critical points since all of them have at least one $0$ entry while $x,y$ and $z$ are given positive.
answered Feb 1 at 4:10
RhaldrynRhaldryn
352416
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By definition, $lambda$ is a constant, and we don't even really care what $lambda$ is. It is always linear and factored from the gradient, so instead of finding $lambda$, use it to jump from one equation to another.
$lambda=yz^2$
$lambda=frac{xz^2}{2}$
$lambda=frac{2xyz}{3}$
answered May 1 '18 at 6:14
AEngineerAEngineer
1,6521318
$endgroup$
1
$begingroup$
by equating values of lambda and the constraint x + 2y + 3z = 1 the values of x,y,z can be calculated and i got the answer x = 1/4, y= 1/8 and z = 1/6.
$endgroup$
– Priya Wadhwa
May 1 '18 at 6:41
add a comment |
$begingroup$
By definition, $lambda$ is a constant, and we don't even really care what $lambda$ is. It is always linear and factored from the gradient, so instead of finding $lambda$, use it to jump from one equation to another.
$lambda=yz^2$
$lambda=frac{xz^2}{2}$
$lambda=frac{2xyz}{3}$
answered May 1 '18 at 6:14
AEngineerAEngineer
1,6521318
$endgroup$
1
$begingroup$
by equating values of lambda and the constraint x + 2y + 3z = 1 the values of x,y,z can be calculated and i got the answer x = 1/4, y= 1/8 and z = 1/6.
$endgroup$
– Priya Wadhwa
May 1 '18 at 6:41
add a comment |
$begingroup$
By definition, $lambda$ is a constant, and we don't even really care what $lambda$ is. It is always linear and factored from the gradient, so instead of finding $lambda$, use it to jump from one equation to another.
$lambda=yz^2$
$lambda=frac{xz^2}{2}$
$lambda=frac{2xyz}{3}$
answered May 1 '18 at 6:14
AEngineerAEngineer
1,6521318
$endgroup$
By definition, $lambda$ is a constant, and we don't even really care what $lambda$ is. It is always linear and factored from the gradient, so instead of finding $lambda$, use it to jump from one equation to another.
$lambda=yz^2$
$lambda=frac{xz^2}{2}$
$lambda=frac{2xyz}{3}$
answered May 1 '18 at 6:14
AEngineerAEngineer
1,6521318
answered May 1 '18 at 6:14
AEngineerAEngineer
1,6521318
answered May 1 '18 at 6:14
AEngineerAEngineer
1,6521318
answered May 1 '18 at 6:14
AEngineerAEngineer
1,6521318
1,6521318
1
$begingroup$
by equating values of lambda and the constraint x + 2y + 3z = 1 the values of x,y,z can be calculated and i got the answer x = 1/4, y= 1/8 and z = 1/6.
$endgroup$
– Priya Wadhwa
May 1 '18 at 6:41
add a comment |
1
$begingroup$
by equating values of lambda and the constraint x + 2y + 3z = 1 the values of x,y,z can be calculated and i got the answer x = 1/4, y= 1/8 and z = 1/6.
$endgroup$
– Priya Wadhwa
May 1 '18 at 6:41
1
1
$begingroup$
by equating values of lambda and the constraint x + 2y + 3z = 1 the values of x,y,z can be calculated and i got the answer x = 1/4, y= 1/8 and z = 1/6.
$endgroup$
– Priya Wadhwa
May 1 '18 at 6:41
$begingroup$
by equating values of lambda and the constraint x + 2y + 3z = 1 the values of x,y,z can be calculated and i got the answer x = 1/4, y= 1/8 and z = 1/6.
$endgroup$
– Priya Wadhwa
May 1 '18 at 6:41
add a comment |
$begingroup$
Alt. hint (no calculus): the following inequality holds by AM-GM, with equality iff $displaystyle x = 2y = frac{3}{2}z,$:
$$
frac{sqrt{3}}{sqrt[4]{2}} cdot sqrt[4]{xyz^2} = sqrt[4]{x cdot 2y cdot frac{3}{2}z cdot frac{3}{2}z} ;le; frac{x+2y+3z}{4} = frac{1}{4} ;;iff;;xyz^2 le frac{2}{3^2 cdot 4^4} = frac{1}{1152}
$$
answered May 1 '18 at 7:05
dxivdxiv
58k648102
$endgroup$
add a comment |
$begingroup$
Alt. hint (no calculus): the following inequality holds by AM-GM, with equality iff $displaystyle x = 2y = frac{3}{2}z,$:
$$
frac{sqrt{3}}{sqrt[4]{2}} cdot sqrt[4]{xyz^2} = sqrt[4]{x cdot 2y cdot frac{3}{2}z cdot frac{3}{2}z} ;le; frac{x+2y+3z}{4} = frac{1}{4} ;;iff;;xyz^2 le frac{2}{3^2 cdot 4^4} = frac{1}{1152}
$$
answered May 1 '18 at 7:05
dxivdxiv
58k648102
$endgroup$
add a comment |
$begingroup$
Alt. hint (no calculus): the following inequality holds by AM-GM, with equality iff $displaystyle x = 2y = frac{3}{2}z,$:
$$
frac{sqrt{3}}{sqrt[4]{2}} cdot sqrt[4]{xyz^2} = sqrt[4]{x cdot 2y cdot frac{3}{2}z cdot frac{3}{2}z} ;le; frac{x+2y+3z}{4} = frac{1}{4} ;;iff;;xyz^2 le frac{2}{3^2 cdot 4^4} = frac{1}{1152}
$$
answered May 1 '18 at 7:05
dxivdxiv
58k648102
$endgroup$
Alt. hint (no calculus): the following inequality holds by AM-GM, with equality iff $displaystyle x = 2y = frac{3}{2}z,$:
$$
frac{sqrt{3}}{sqrt[4]{2}} cdot sqrt[4]{xyz^2} = sqrt[4]{x cdot 2y cdot frac{3}{2}z cdot frac{3}{2}z} ;le; frac{x+2y+3z}{4} = frac{1}{4} ;;iff;;xyz^2 le frac{2}{3^2 cdot 4^4} = frac{1}{1152}
$$
answered May 1 '18 at 7:05
dxivdxiv
58k648102
answered May 1 '18 at 7:05
dxivdxiv
58k648102
answered May 1 '18 at 7:05
dxivdxiv
58k648102
answered May 1 '18 at 7:05
dxivdxiv
58k648102
58k648102
add a comment |
add a comment |
$begingroup$
Hint: prove that $$xyz^2le frac{1}{1152}$$ and the equal sign holds if $$x=frac{1}{4},y=frac{1}{8},z=frac{1}{6}$$
answered May 1 '18 at 5:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
$endgroup$
add a comment |
$begingroup$
Hint: prove that $$xyz^2le frac{1}{1152}$$ and the equal sign holds if $$x=frac{1}{4},y=frac{1}{8},z=frac{1}{6}$$
answered May 1 '18 at 5:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
$endgroup$
add a comment |
$begingroup$
Hint: prove that $$xyz^2le frac{1}{1152}$$ and the equal sign holds if $$x=frac{1}{4},y=frac{1}{8},z=frac{1}{6}$$
answered May 1 '18 at 5:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
$endgroup$
Hint: prove that $$xyz^2le frac{1}{1152}$$ and the equal sign holds if $$x=frac{1}{4},y=frac{1}{8},z=frac{1}{6}$$
answered May 1 '18 at 5:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
answered May 1 '18 at 5:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
answered May 1 '18 at 5:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
answered May 1 '18 at 5:53
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.8k42867
78.8k42867
add a comment |
add a comment |
$begingroup$
You can also convert $xyz^2$ into a function of two variables $f(y,z)=(1-2y-3z)yz^2$ and use this test to conclude that the maxima occurs at $(frac{1}{4},frac{1}{8}, frac{1}{6})$ and is $1152.$ You can ignore the other critical points since all of them have at least one $0$ entry while $x,y$ and $z$ are given positive.
answered Feb 1 at 4:10
RhaldrynRhaldryn
352416
$endgroup$
add a comment |
$begingroup$
You can also convert $xyz^2$ into a function of two variables $f(y,z)=(1-2y-3z)yz^2$ and use this test to conclude that the maxima occurs at $(frac{1}{4},frac{1}{8}, frac{1}{6})$ and is $1152.$ You can ignore the other critical points since all of them have at least one $0$ entry while $x,y$ and $z$ are given positive.
answered Feb 1 at 4:10
RhaldrynRhaldryn
352416
$endgroup$
add a comment |
$begingroup$
You can also convert $xyz^2$ into a function of two variables $f(y,z)=(1-2y-3z)yz^2$ and use this test to conclude that the maxima occurs at $(frac{1}{4},frac{1}{8}, frac{1}{6})$ and is $1152.$ You can ignore the other critical points since all of them have at least one $0$ entry while $x,y$ and $z$ are given positive.
answered Feb 1 at 4:10
RhaldrynRhaldryn
352416
$endgroup$
You can also convert $xyz^2$ into a function of two variables $f(y,z)=(1-2y-3z)yz^2$ and use this test to conclude that the maxima occurs at $(frac{1}{4},frac{1}{8}, frac{1}{6})$ and is $1152.$ You can ignore the other critical points since all of them have at least one $0$ entry while $x,y$ and $z$ are given positive.
answered Feb 1 at 4:10
RhaldrynRhaldryn
352416
answered Feb 1 at 4:10
RhaldrynRhaldryn
352416
answered Feb 1 at 4:10
RhaldrynRhaldryn
352416
answered Feb 1 at 4:10
RhaldrynRhaldryn
352416
352416
add a comment |
add a comment |
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