An open box with square base is to be constructed. The material for the base costs $10 per square foot. The material for the sides costs $1 per square foot. the box must have an area of 100 square feet. Find the dimensions of the box that minimize cost.

What I have so far:

A = x^2 + 4hx = 100
V = x^2 * h

I solved for H in the area formula to get:
h= x^2(100-x^2/4x)
I then put this into the volume formula; however, i'm lost from there.

Any help would be greatly appreciated.

Cheers,

c = 10 x^2 + 4 x h

but
h = (100 - x^2)/4x

c = 10 x^2 + 4x (100-x^2)/4x

c = 10 x^2 + 100 - x^2

c = 9 x^2 + 100

cost is minimum as x goes to zero of course since the bottom material is so expensive. I have a hunch that the constraint should be on the volume, not the area.

for example see

http://www.jiskha.com/display.cgi?id=1289778468

Hmm, even so though, it doesn't make too much sense.

V=x^2h = 100
h= 100/x^2
C = x^2 + 4xh

C= 10x^2+4(1)xh

C= 10x^2+4x(100/x^2)
10x^2+400x/4x^2
C= 10x^2 + 100/x

C'(x) = 20x-100/x^2

20x^3=100

x^3 = 5

5^(1/3) = 1.7099
y = 100/5^(1/3)

y = 58.5

That seems like a lot for a minimum.

To find the dimensions of the box that minimize cost, you need to express the cost function in terms of a single variable and then find its minimum. Let's call the side length of the square base x and the height of the box h.

The cost of the base material is $10 per square foot, so the cost of the base is 10x^2 dollars.

The cost of the side material is $1 per square foot, and the box has four sides, so the cost of the sides is 4xh dollars.

The total cost, C, is the sum of the cost of the base and the cost of the sides:
C = 10x^2 + 4xh.

To eliminate h, you can use the volume formula: V = x^2 * h = 100. Rearranging this equation, you get h = 100 / x^2.

Now substitute this expression for h back into the cost function:
C = 10x^2 + 4x * (100 / x^2) = 10x^2 + 400 / x.

To find the dimensions of the box that minimize the cost, you need to find the value of x that minimizes the cost function C.

Taking the derivative of C with respect to x and setting it equal to 0 will give you the critical points:
dC/dx = 20x - 400 / x^2 = 0.
Simplifying, you get:
20x = 400 / x^2
20x^3 = 400
x^3 = 20
x = cuberoot(20)

Now, you need to check if this critical point x = cuberoot(20) is indeed the minimum. To do so, you can take the second derivative of C and evaluate it at this critical point.

Taking the second derivative of C with respect to x:
d^2C/dx^2 = 20 + 800 / x^3.

Plugging in the critical point x = cuberoot(20):
d^2C/dx^2 = 20 + 800 / (cuberoot(20))^3
= 20 + 800 / 20
= 20 + 40
= 60.

Since the second derivative is positive (60), the critical point x = cuberoot(20) corresponds to a minimum.

Therefore, the dimensions of the box that minimize cost are:
- Side length of the square base, x = cuberoot(20)
- Height of the box, h = 100 / (cuberoot(20))^2.