Volume and Area of Rectangular Container

[mathnovice11]

A company is designing a rectangular container with a square base (with a side length of x). The company wants the container to hold a volume of 64 cubic inches and they would like to minimize the amount of material used for the container. What are the dimensions of the container, and what is the minimum amount of material needed. If the height of the container is given by h, the volume of the container is V=x^2*h and the surface area of the container is S=2x^2+4xh. Kepp at least 4 decimal places in your computed values for the side of the base and the height and give the minimum surface area rounded to two decimal places. Show all your work, including a first derivative chart that allows you to conclude that you have found the desired minimum

I have the following work done: base length=x
height= h
Volume= x^2*h
64=x^2*h

S=2x^2+4xh
h= 64/x^2 - 1
s= 2x^2 + 4x(64/x^2)

 

[galactus]

You almost have it.

\(\displaystyle V=x^{2}h=64\)

\(\displaystyle S=2x^{2}+4xh\)

Since we are minimizing S, we solve V for h and sub into S. If we were minimizing V, we would solve S for h and sub into V.

\(\displaystyle x^{2}h=64 \Rightarrow h=\frac{64}{x^{2}}\)...............[1]

\(\displaystyle S=2x^{2}+4x(\frac{64}{x^{2}})=2x^{2}+\frac{256}{x}\)

Now, S is entriely in terms of one variable, x.

Differentiate: \(\displaystyle \frac{dS}{dx}=4x-\frac{256}{x^{2}}\)

Set to 0 and solve for x. Then, h can be found by subbing this value back into [1].

Notice, that the surface area is minimized when the box is a cube. The dimensions are integers, so no rounding necessary.