# Homework Help: Optimization: Minimum Surface Area

1. Jul 19, 2008

### JoeyC2488

1. The problem statement, all variables and given/known data

I need help on an optimization problem involving a hexagonal prism with no bottom or top, but the top is covered by a trihedral pyramid which has a displacement, x, such that the surface area of the object is at a minimum for a given volume. The assigned variables include:
A= the surface area of the object
a= the length of each side of the hexagon (a=1)
h= the height of the prism
x= displacement

2. Relevant equations

Given that A= 6(ah-1/2ax) + 3a$$\sqrt{3}$$*$$\sqrt{x^2+(a^2/4)}$$, use calculus to find the displacement, x that yields the minimum surface area. Calculate x if a=1.

3. The attempt at a solution

I'm struggling trying to find a secondary equation for this optimization problem and I think I need to use an equation evolving h, but I'm not sure what.

2. Jul 19, 2008

### arildno

The volume V is to be constant, this will lead you to an equation in h and x, from which you may find an expression for h.

3. Jul 20, 2008

### JoeyC2488

How would I use a formula for volume since the problem works with two shapes.

V=(3$$\sqrt{3}$$)/2 a^2*h

h= V/ (3$$\sqrt{3}$$/2 a^2*h

Then, I can just plug this into the first equation, simplify, and start the calculus. Is that right?

4. Jul 20, 2008

### arildno

Your TOTAL volume is the sum of the prism's volume and the pyramid's volume.

That is the equation in h and x you are to set up!

5. Jul 20, 2008

### arildno

$$V=\frac{3\sqrt{3}}{2}a^{2}h+\frac{1}{3}\frac{3\sqrt{3}}{2}a^{2}x\to{h}=\frac{2V}{3\sqrt{3}a^{2}}-\frac{x}{3}$$

6. Jul 20, 2008

### JoeyC2488

So after substituting that in for h, the simplified form would be:

A= 4V-5x$$\sqrt{3}$$+ 9$$\sqrt{x^2+4}$$

$$\sqrt{3}$$

7. Jul 20, 2008

### JoeyC2488

How would I take the derivative of this equation then with multiple variables?

8. Jul 21, 2008

### arildno

You only have one variable, x. Remember, V is a constant!