# Finding global minimum.

• November 17th 2009, 06:05 PM
Rumor
Finding global minimum.
At least, I think that's what this question is asking for.

"The total cost to build an open box with a constant volume V and with a square base is given by the function

$C(x)=10x^2+\frac{60V}{x}$

where x is the length of the square base.

a) Find the value of x that will produce the box with the minimum total cost. Your answer will involve the parameter of V.

b) Verify that this value of x will give the local minimum.

c) Give an argument based on calculus why this value of x actually gives a global minimum, not just a local minimum."

I'm not sure how to start.
• November 17th 2009, 06:22 PM
RockHard
I know I won't be of some much help since I was never the best at these types of problems but usually finding a function say $f(x)$ maxima and minima you can use the method of using the first derivative and solving for x which will give you critical points on the graph and then plug in the critical points you found in the original function then it should not be hard to determine which is the maxima and minima also don't forget to test the end points of the graph if they are included which means the values of x which the graph is contained which can similar be used in part for you answer in part 3, also it is quite simple the graph will be contained on some value greater than 0 because It should not be negative which in reality is just like saying someone is paying you to make a box but in this case your making it yourself
• November 17th 2009, 06:22 PM
mr fantastic
Quote:

Originally Posted by Rumor
At least, I think that's what this question is asking for.

"The total cost to build an open box with a constant volume V and with a square base is given by the function

$C(x)=10x^2+\frac{60V}{x}$

where x is the length of the square base.

a) Find the value of x that will produce the box with the minimum total cost. Your answer will involve the parameter of V.

b) Verify that this value of x will give the local minimum.

c) Give an argument based on calculus why this value of x actually gives a global minimum, not just a local minimum."

I'm not sure how to start.

Start by solving dC/dx = 0. If you need more help, please show all your work and say where you're stuck.
• November 17th 2009, 07:55 PM
Rumor
Hm. I feel as if I'm doing something wrong when taking the derivative.

$10x^2$ turns into $20x$.

And for 60/x, the quotient rule comes into play, so it's (60*x)-(60V*1) all over $x^2$

Therefore,

$C'(x)=20x+\frac{(60x-60V)}{x^2}$

Is this right?
• November 17th 2009, 08:25 PM
RockHard
Hint:

You can write $\frac{60V}{x}$ as

$60V{x^{-1}}$
• November 18th 2009, 02:45 AM
mr fantastic
Quote:

Originally Posted by Rumor
Hm. I feel as if I'm doing something wrong when taking the derivative.

$10x^2$ turns into $20x$.

And for 60/x, the quotient rule comes into play, so it's (60*x)-(60V*1) all over $x^2$

Therefore,

$C'(x)=20x+\frac{(60x-60V)}{x^2}$

Is this right?

No. Note that the derivative of a constant is (eg. V) is zero.
• November 18th 2009, 03:10 AM
simplependulum
The greatest method is using $AM \geq GM$

$C = 10x^2 + \frac{60V}{x} = 10x^2 + \frac{30V}{x} + \frac{30V}{x} \geq 3 \cdot \left ( \frac{10x^2 \cdot 30V \cdot 30V }{x ^2}\right ) ^{\frac{1}{3}}$

$= 30 ( 9V^2 )^{\frac{1}{3}}$ min. value

when $\frac{30V}{x} = 10x^2 \implies x = (3V)^{\frac{1}{3}}$
• November 18th 2009, 10:33 AM
RockHard
Good example SP, I got a similar answer