A manufacturer wants to design an open box (no top) having a square base and
a fixed volume. To reduce the cost the manufacturer requires the box to enclose a volume of 108 in^3 using the least amount of material. What dimensions for the box will minimize its surface area?
maximum volume?

Question

A manufacturer wants to design an open box (no top) having a square base and
a fixed volume. To reduce the cost the manufacturer requires the box to enclose a volume of 108 in^3 using the least amount of material. What dimensions for the box will minimize its surface area?
maximum volume?

kirbykirby · Answer

Yes sorry I just jotting stuff on paper

anonymous · Answer

oh ok its fine

kirbykirby · Answer

First, let's look at how we can maximizae the volume:

Ok So what you can do is establish the formula for the surface area of the box... Let $c$ denote the side of the square base. The other 4 sides will be rectangles (unless your maximization happens to give square sides).

Thus you get $A=c^2+4(cL)$, where $L$ is the length of the side of the rectangle.

You can write down the volume of the Box to be$V=c^2*L=108$

Ok so far?

kirbykirby · Answer

Wait actually I just realized. they want to maximize the volume? But the volume is given (in my head I thought we were maximizing the surface area)

kirbykirby · Answer

Are they only just interested in minimizing the surface area then?

kirbykirby · Answer

Because the idea in these problems is to always express the function you are trying to maximize/minimize in terms of one variable. Then  take the derivative of the function you're maximizing/minimizing and set it equal to 0 and solve for the variable, and perform a 2nd derivative test to see which point(s) are maxima or minima

kirbykirby · Answer

Anyway, if we are minimizing the surface area... You can solve for $L$ easily in the volume equation:

$$c^2L=108\
L=\frac{108}{c^2}$$

You can now substitute $L$ in the formula for the Area using the above:

$$A=c^2+4c\left(\frac{108}{c^2}\right)$$

THe idea is that you are given a volume of 108 that must be fixed. The area of the box is what is going to vary because that's what we want to minimize. As you know, the 1st derivative in calculus is used to find maximums and minimums. 
So, we will do the 1st derivative test on the equation of the Area to find its extrema. But, the area is expressed as a function of 2 variables, that's why we used the equation of the volume to make $L$ in terms of $c$ only ...so that's why you use the equation of the volume to do this.

Now the area is a function of $c$ only, so you take the derivative:
$$\frac{dA}{dc}=\frac{d}{dc}\left( c^2+4c \left(\frac{108}{c^2}\right) \right) $$

kirbykirby · Answer

$$\frac{dA}{dc}=\frac{d}{dc}\left( c^2+4c \left(\frac{108}{c^2}\right) \right)\
=\frac{d}{dc}\left( c^2+\frac{432}{c}\right)\
=2c-\frac{432}{c^2} $$

Now you set this equal to 0:
$$2c-\frac{432}{c^2}=0\
2c^3-432=0\
c^3-216=0\
c^3=216\
c=6$$

Now you verify that this is a minimum using the second derivative ->take the 2nd derivative of the Area equation:
$$\frac{d^2A}{dc^2}=\frac{d}{dc}\left(\frac{dA}{dc}\right)=\frac{d}{dc}\left( 2c -\frac{432}{c^2}  \right)\
=2+(2)\frac{432}{c^3}$$

If you lug in $c=6)$ in that last expression above, you'll see that it is a positive value (2nd derivative is positive), implying that we have a minimum by the 2nd derivative test.

kirbykirby · Answer

plug in c=6 *

kirbykirby · Answer

I really don't know about the maximization part of the question because they already set a volume as your constraint..

anonymous · Answer

but why the dA/dC?

kirbykirby · Answer

You want to optimize the Area of the box. You find min/max of function by finding the 1st derivative and setting them to 0 (1st derivative test). We are doing the same thing here, by finding the derivative of the Area function and setting it to 0