Find maximum volume

Question

A 160-inch strip of metal 20 inches wide is to be made into a small open trough by bending up two sides on the long side, at right angles to the base. The sides will be the same height, x. If the trough is to have a maximum volume, how many inches should be turned up on each side? 

Answer 1

It is unclear what the shape of the trough is by reading the problem. Although a trough is often triangular in cross-section, there is no reason that it couldn't be rectangular in cross-section. I'll pursue an answer assuming it is rectangular.

The volume of this trough is given by:

V(x) = (160 - 2x)(20 - 2x)(x)

V(x) = (3200 - 360x + 4x²)(x)

V(x) = 3200x - 360x² + 4x³

To find the maximum, you can either graph this and look for the maximum or you can take the derivative and set it equal to zero.

V'(x) = 3200 - 720x + 12x²

3200 - 720x + 12x² = 0

3x² - 180x + 800 = 0

Using the quadratic formula, x = 4.834 and x = 55.166

We can throw out the second solution because the width is only 20 (in reality, this solution is a minimum value mathematically anyway).

So 4.834 inches should be turned up on each side.

Answer 2

Your triangular trough at 160, 10, 10, at a rt angle has a volume of:

side area=ab/2=50*160=8000cu/in.

Square=hwl

x+2y=20

18*1=18sq/in

16*2=32 "

14*3=42 "

12*4=48 "

10*5=50 " *****************

8*6=48 "

Answer 3

The trough will ultimately have the shape of a long rectangular prism (with no top, but the volume is the same). So to calculate volume, you will need to know the length, the height, and the width.

The length is given: 160 inches.

The height is unknown, but is the same on both sides: say, x inches

The width was 20 inches, but was bent to form two sides of height x: 20-2x inches.

Thus, the volume is 160*x*(20-2x) in^3. That is, as a function of x, the volume V(x) is given by:

V(x)=-320x^2+3200x.

To find the maximum volume, you need to know where the critical points of this function (an upside down parabola) are. This is where the derivative is equal to 0. Thus, we want to solve

V'(x)=-640x+3200 = 0

From which we find that x=5. Thus, the volume is maximized when the height x=5. At this point, the volume is V(5)= -320(5)^2 + 3200(5)=8000 in^3.