Help with calculus please!

Question

I need help with this. A 20-inch wide of 160 inches strip of metal will be made into an open trough by bending up two sides on the long side based on the right angles to the base. Furthermore, the two sides will have the same height, x. How many inches should be turned up on each side if the trough is to have a maximum volume?

Answer 1

V = (20 - 2x)(160)

V = -320x² + 3200x

dV/dx = -640x + 3200

0 = -640(x - 5)

Maximum at x = 5.

Answer 2

I am assuming that strip of metal is 20 inches wide and 160 inches long. Initially, the strip is flat. Two sides of length x will be turned up on the long end so 2x must be subtracted from 160. Now, the dimensions will now be 20 inches by (160 - 2x) inches. This will give us a height of x inches as mentioned in the question. We can understand that this figure is similar to a box. So the volume will be length x width x height. In our case, this will be 20 * (160 -2x) * x. Therefore, the volume function is V = 3200x-40x^2.

To find the maximum volume, we need to

a.) Find the derivative.

b) Set it to 0 to find x.

c) Plug that x into V.

The derivative of this function gives us 3200-80x. Setting this to 0, we get x=40. Plugging x into V, we get 3200(40) - 40(40)^2 = 64000. So that maximum volume will be 64000 inches cubed.

Hope I was able to help! Feel free to reach out with further questions!