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March 4, 2015

March 4, 2015

Posted by **Ann** on Monday, February 17, 2014 at 10:47pm.

The answer is 5 X 5 X 2, but I don't understand how to get that.

Assuming that length = x, width = x. height = h, these are the two equations I came up with:

(x^2)h = 50

SA = x^2 + 4xh

Am I doing this right?

- Calculus -
**Reiny**, Monday, February 17, 2014 at 11:48pmYou are doing fine so far.

For these kind of max/min questions, look for "something" which will be either maximized or minimized. What ever that "something" is , you will need an equation that says

"something" = ......

In this case I see, " .... what dimensions will minimize the**cost**of construction? "

So in this case it is the COST

cost = 4(x^2) + 5(4xh)

= 4x^2 + 20xh

but we know:

x^2 h = 50

so h = 50/x^2

cost = 4x^2 + 20x(50/x^2)

= 4x^2 + 1000/x

now differentiate with respect to x

d(cost)/dx= 8x - 1000/x^2

= 0 for a min of cost

8x = 1000/x^2

8x^3 = 1000

take cube root of both sides

2x = 10

x = 5

then h = 50/5^2 = 2

and there you have it!

base is 5 by 5 , and the height is 2

- Calculus -
**Ann**, Tuesday, February 18, 2014 at 12:02amThanks so much Reiny!

I never thought of adding the cost to the equation, but now I know.

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