1) A box is to be made out of 12 square feet of material. The marketing department wants the length of the box to be twice its width. What dimensions will maximize the volume of the box?

I have the obj function as $\displaystyle V (l, w, h) = lwh $ and since $\displaystyle l = 2w $ ... $\displaystyle V (w, h) = 2w^{2}h $

I believe the next step is to calculate the area... $\displaystyle A = lw $ ... $\displaystyle 12 ft^{2} = lw $ ... solving for w gets $\displaystyle \frac{12 ft^{2}}{l} = w $

Pluggin the constraint into the obj function I get $\displaystyle V(h) = 2(\frac{12 ft^{2}}{l})^{2}h $ , and I'm think I'm heading in the wrong direction because I end up with $\displaystyle ft^{4} $ ...

2) 3 meters of wire is supposed to be cut into two pieces. Then, one piece will be reshaped as a square, and the other into a circle. Find the lengths of each piece to minimize the sum of the areas of the resulting shapes.

________________= 3 meters long and broken into two parts 'a' and 3 - 'a'

| 'a' | 'b' or 3 - 'a' |

The circle has $\displaystyle Circumference = 2{\pi}r $ ... Using 'a' as the length of the circumference $\displaystyle a = 2{\pi}r $ or $\displaystyle r = \frac{a}{2\pi} $ and plugging that into the equation for area of a circle $\displaystyle A = \pi({\frac{a}{2\pi}})^{2} = \frac{a^{2}}{4{\pi}}$

For the square I have $\displaystyle \frac{3-a}{4} $ as the lengths of each side. Thus, the area $\displaystyle A = (\frac{3-a}{4})^{2} $ which factors out to $\displaystyle \frac{a^{2}-6a+9}{16} $

If I calculated the areas for the respective shapes correctly, do I just take the derivative and solve for 'a' at this point?

Thanks!