Mixing
Solving the minimum value for resistance
$begingroup$
How to solve the problem by using calculus- local maxima and minima ?(By first derivative test)
I solved it but I can't find the local maximum and minimum points.
algebra-precalculus
asked Jan 5 at 16:29
user218102user218102
234
$endgroup$
add a comment |
$begingroup$
How to solve the problem by using calculus- local maxima and minima ?(By first derivative test)
I solved it but I can't find the local maximum and minimum points.
algebra-precalculus
asked Jan 5 at 16:29
user218102user218102
234
$endgroup$
$begingroup$
Maybe you should post your progress so far, so we can give you better feedback.
$endgroup$
– 0x539
Jan 5 at 16:37
$begingroup$
I posted up there.
$endgroup$
– user218102
Jan 5 at 16:49
$begingroup$
Your link is asking for a Google sign-in.
$endgroup$
– 0x539
Jan 5 at 16:50
$begingroup$
What about now?
$endgroup$
– user218102
Jan 5 at 16:54
$begingroup$
I changed it .Thanks for your response .
$endgroup$
– user218102
Jan 5 at 16:58
add a comment |
$begingroup$
How to solve the problem by using calculus- local maxima and minima ?(By first derivative test)
I solved it but I can't find the local maximum and minimum points.
algebra-precalculus
asked Jan 5 at 16:29
user218102user218102
234
$endgroup$
How to solve the problem by using calculus- local maxima and minima ?(By first derivative test)
I solved it but I can't find the local maximum and minimum points.
algebra-precalculus
algebra-precalculus
asked Jan 5 at 16:29
user218102user218102
234
asked Jan 5 at 16:29
user218102user218102
234
edited Jan 5 at 16:57
user218102
asked Jan 5 at 16:29
user218102user218102
234
asked Jan 5 at 16:29
user218102user218102
234
asked Jan 5 at 16:29
user218102user218102
234
234
$begingroup$
Maybe you should post your progress so far, so we can give you better feedback.
$endgroup$
– 0x539
Jan 5 at 16:37
$begingroup$
I posted up there.
$endgroup$
– user218102
Jan 5 at 16:49
$begingroup$
Your link is asking for a Google sign-in.
$endgroup$
– 0x539
Jan 5 at 16:50
$begingroup$
What about now?
$endgroup$
– user218102
Jan 5 at 16:54
$begingroup$
I changed it .Thanks for your response .
$endgroup$
– user218102
Jan 5 at 16:58
add a comment |
$begingroup$
Maybe you should post your progress so far, so we can give you better feedback.
$endgroup$
– 0x539
Jan 5 at 16:37
$begingroup$
I posted up there.
$endgroup$
– user218102
Jan 5 at 16:49
$begingroup$
Your link is asking for a Google sign-in.
$endgroup$
– 0x539
Jan 5 at 16:50
$begingroup$
What about now?
$endgroup$
– user218102
Jan 5 at 16:54
$begingroup$
I changed it .Thanks for your response .
$endgroup$
– user218102
Jan 5 at 16:58
$begingroup$
Maybe you should post your progress so far, so we can give you better feedback.
$endgroup$
– 0x539
Jan 5 at 16:37
$begingroup$
Maybe you should post your progress so far, so we can give you better feedback.
$endgroup$
– 0x539
Jan 5 at 16:37
$begingroup$
I posted up there.
$endgroup$
– user218102
Jan 5 at 16:49
$begingroup$
I posted up there.
$endgroup$
– user218102
Jan 5 at 16:49
$begingroup$
Your link is asking for a Google sign-in.
$endgroup$
– 0x539
Jan 5 at 16:50
$begingroup$
Your link is asking for a Google sign-in.
$endgroup$
– 0x539
Jan 5 at 16:50
$begingroup$
What about now?
$endgroup$
– user218102
Jan 5 at 16:54
$begingroup$
What about now?
$endgroup$
– user218102
Jan 5 at 16:54
$begingroup$
I changed it .Thanks for your response .
$endgroup$
– user218102
Jan 5 at 16:58
$begingroup$
I changed it .Thanks for your response .
$endgroup$
– user218102
Jan 5 at 16:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$F'(x) > 0$ is wrong for the interval $left( -frac1{sqrt{5}}, frac1{sqrt{5}}right)$. Try plugging in $x = frac1{10}$ for example.
answered Jan 5 at 17:02
0x5390x539
1,271317
$endgroup$
add a comment |
$begingroup$
Method $1$
As you have mentioned, by setting the 1st-order derivative equal to zero we attain to two points $$x=pm{1over 2sqrt 5}$$ Also $$f''(x)={10over x^3}$$since for $x={1over 2sqrt 5}$ we have $f''(x)>0$ therefore the point $left({1over 2sqrt 5},20sqrt 5right)$ is a local minimum. Similarly the point $left(-{1over 2sqrt 5},-20sqrt 5right)$ is a local maximum.
Method $2$
We have $$f(x)={5over x}+100x=10sqrt 5left({1over 2sqrt 5x}+2sqrt 5xright)ge 20sqrt 5$$where the equality happens if and only if $x={1over 2sqrt 5}$. Similarly for $x=-{1over 2sqrt 5}$
answered Jan 5 at 17:07
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
$begingroup$
Why we get a local maximum point when f ''(X)<0 ?
$endgroup$
– user218102
Jan 5 at 17:14
$begingroup$
In fact this follows directly from the Taylor expansion:$$f(x+x_0)=f(x)+f'(x)cdot x_0+{1over 2}x_0^2cdot f''(x)+o(x_0^2)$$where $o(.)$ is the famous little-o notation. Also we should have $f'(x)=0$ in local maximum to conclude that.
$endgroup$
– Mostafa Ayaz
Jan 5 at 17:18
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
$F'(x) > 0$ is wrong for the interval $left( -frac1{sqrt{5}}, frac1{sqrt{5}}right)$. Try plugging in $x = frac1{10}$ for example.
answered Jan 5 at 17:02
0x5390x539
1,271317
$endgroup$
add a comment |
$begingroup$
$F'(x) > 0$ is wrong for the interval $left( -frac1{sqrt{5}}, frac1{sqrt{5}}right)$. Try plugging in $x = frac1{10}$ for example.
answered Jan 5 at 17:02
0x5390x539
1,271317
$endgroup$
add a comment |
$begingroup$
$F'(x) > 0$ is wrong for the interval $left( -frac1{sqrt{5}}, frac1{sqrt{5}}right)$. Try plugging in $x = frac1{10}$ for example.
answered Jan 5 at 17:02
0x5390x539
1,271317
$endgroup$
$F'(x) > 0$ is wrong for the interval $left( -frac1{sqrt{5}}, frac1{sqrt{5}}right)$. Try plugging in $x = frac1{10}$ for example.
answered Jan 5 at 17:02
0x5390x539
1,271317
answered Jan 5 at 17:02
0x5390x539
1,271317
answered Jan 5 at 17:02
0x5390x539
1,271317
answered Jan 5 at 17:02
0x5390x539
1,271317
1,271317
add a comment |
add a comment |
$begingroup$
Method $1$
As you have mentioned, by setting the 1st-order derivative equal to zero we attain to two points $$x=pm{1over 2sqrt 5}$$ Also $$f''(x)={10over x^3}$$since for $x={1over 2sqrt 5}$ we have $f''(x)>0$ therefore the point $left({1over 2sqrt 5},20sqrt 5right)$ is a local minimum. Similarly the point $left(-{1over 2sqrt 5},-20sqrt 5right)$ is a local maximum.
Method $2$
We have $$f(x)={5over x}+100x=10sqrt 5left({1over 2sqrt 5x}+2sqrt 5xright)ge 20sqrt 5$$where the equality happens if and only if $x={1over 2sqrt 5}$. Similarly for $x=-{1over 2sqrt 5}$
answered Jan 5 at 17:07
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
$begingroup$
Why we get a local maximum point when f ''(X)<0 ?
$endgroup$
– user218102
Jan 5 at 17:14
$begingroup$
In fact this follows directly from the Taylor expansion:$$f(x+x_0)=f(x)+f'(x)cdot x_0+{1over 2}x_0^2cdot f''(x)+o(x_0^2)$$where $o(.)$ is the famous little-o notation. Also we should have $f'(x)=0$ in local maximum to conclude that.
$endgroup$
– Mostafa Ayaz
Jan 5 at 17:18
add a comment |
$begingroup$
Method $1$
As you have mentioned, by setting the 1st-order derivative equal to zero we attain to two points $$x=pm{1over 2sqrt 5}$$ Also $$f''(x)={10over x^3}$$since for $x={1over 2sqrt 5}$ we have $f''(x)>0$ therefore the point $left({1over 2sqrt 5},20sqrt 5right)$ is a local minimum. Similarly the point $left(-{1over 2sqrt 5},-20sqrt 5right)$ is a local maximum.
Method $2$
We have $$f(x)={5over x}+100x=10sqrt 5left({1over 2sqrt 5x}+2sqrt 5xright)ge 20sqrt 5$$where the equality happens if and only if $x={1over 2sqrt 5}$. Similarly for $x=-{1over 2sqrt 5}$
answered Jan 5 at 17:07
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
$begingroup$
Why we get a local maximum point when f ''(X)<0 ?
$endgroup$
– user218102
Jan 5 at 17:14
$begingroup$
In fact this follows directly from the Taylor expansion:$$f(x+x_0)=f(x)+f'(x)cdot x_0+{1over 2}x_0^2cdot f''(x)+o(x_0^2)$$where $o(.)$ is the famous little-o notation. Also we should have $f'(x)=0$ in local maximum to conclude that.
$endgroup$
– Mostafa Ayaz
Jan 5 at 17:18
add a comment |
$begingroup$
Method $1$
As you have mentioned, by setting the 1st-order derivative equal to zero we attain to two points $$x=pm{1over 2sqrt 5}$$ Also $$f''(x)={10over x^3}$$since for $x={1over 2sqrt 5}$ we have $f''(x)>0$ therefore the point $left({1over 2sqrt 5},20sqrt 5right)$ is a local minimum. Similarly the point $left(-{1over 2sqrt 5},-20sqrt 5right)$ is a local maximum.
Method $2$
We have $$f(x)={5over x}+100x=10sqrt 5left({1over 2sqrt 5x}+2sqrt 5xright)ge 20sqrt 5$$where the equality happens if and only if $x={1over 2sqrt 5}$. Similarly for $x=-{1over 2sqrt 5}$
answered Jan 5 at 17:07
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
Method $1$
As you have mentioned, by setting the 1st-order derivative equal to zero we attain to two points $$x=pm{1over 2sqrt 5}$$ Also $$f''(x)={10over x^3}$$since for $x={1over 2sqrt 5}$ we have $f''(x)>0$ therefore the point $left({1over 2sqrt 5},20sqrt 5right)$ is a local minimum. Similarly the point $left(-{1over 2sqrt 5},-20sqrt 5right)$ is a local maximum.
Method $2$
We have $$f(x)={5over x}+100x=10sqrt 5left({1over 2sqrt 5x}+2sqrt 5xright)ge 20sqrt 5$$where the equality happens if and only if $x={1over 2sqrt 5}$. Similarly for $x=-{1over 2sqrt 5}$
answered Jan 5 at 17:07
Mostafa AyazMostafa Ayaz
15.3k3939
answered Jan 5 at 17:07
Mostafa AyazMostafa Ayaz
15.3k3939
answered Jan 5 at 17:07
Mostafa AyazMostafa Ayaz
15.3k3939
answered Jan 5 at 17:07
Mostafa AyazMostafa Ayaz
15.3k3939
15.3k3939
$begingroup$
Why we get a local maximum point when f ''(X)<0 ?
$endgroup$
– user218102
Jan 5 at 17:14
$begingroup$
In fact this follows directly from the Taylor expansion:$$f(x+x_0)=f(x)+f'(x)cdot x_0+{1over 2}x_0^2cdot f''(x)+o(x_0^2)$$where $o(.)$ is the famous little-o notation. Also we should have $f'(x)=0$ in local maximum to conclude that.
$endgroup$
– Mostafa Ayaz
Jan 5 at 17:18
add a comment |
$begingroup$
Why we get a local maximum point when f ''(X)<0 ?
$endgroup$
– user218102
Jan 5 at 17:14
$begingroup$
In fact this follows directly from the Taylor expansion:$$f(x+x_0)=f(x)+f'(x)cdot x_0+{1over 2}x_0^2cdot f''(x)+o(x_0^2)$$where $o(.)$ is the famous little-o notation. Also we should have $f'(x)=0$ in local maximum to conclude that.
$endgroup$
– Mostafa Ayaz
Jan 5 at 17:18
$begingroup$
Why we get a local maximum point when f ''(X)<0 ?
$endgroup$
– user218102
Jan 5 at 17:14
$begingroup$
Why we get a local maximum point when f ''(X)<0 ?
$endgroup$
– user218102
Jan 5 at 17:14
$begingroup$
In fact this follows directly from the Taylor expansion:$$f(x+x_0)=f(x)+f'(x)cdot x_0+{1over 2}x_0^2cdot f''(x)+o(x_0^2)$$where $o(.)$ is the famous little-o notation. Also we should have $f'(x)=0$ in local maximum to conclude that.
$endgroup$
– Mostafa Ayaz
Jan 5 at 17:18
$begingroup$
In fact this follows directly from the Taylor expansion:$$f(x+x_0)=f(x)+f'(x)cdot x_0+{1over 2}x_0^2cdot f''(x)+o(x_0^2)$$where $o(.)$ is the famous little-o notation. Also we should have $f'(x)=0$ in local maximum to conclude that.
$endgroup$
– Mostafa Ayaz
Jan 5 at 17:18
add a comment |
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$begingroup$
Maybe you should post your progress so far, so we can give you better feedback.
$endgroup$
– 0x539
Jan 5 at 16:37
$begingroup$
I posted up there.
$endgroup$
– user218102
Jan 5 at 16:49
$begingroup$
Your link is asking for a Google sign-in.
$endgroup$
– 0x539
Jan 5 at 16:50
$begingroup$
What about now?
$endgroup$
– user218102
Jan 5 at 16:54
$begingroup$
I changed it .Thanks for your response .
$endgroup$
– user218102
Jan 5 at 16:58