Re: 2 Optimization Problems

In the first problem, you are equating the area of the sheet of material, rather than the surface area of the box to the given area. The length and width of the sheet are not the same as the length and width of the box. You want to state:

$\displaystyle 2lw+2hw+2hl=12\text{ ft}^2$

Now use the condition on the length and the above constraint to get the objective function in one variable.

For the second problem, you are doing well...write the total area as a function of $\displaystyle a$, then equate the first derivative to zero to find the critical value, and I suggest that you use the second derivative test to show this critical value minimizes the area.

Re: 2 Optimization Problems

For the first problem I used the correct area $\displaystyle 2lw + 2hw + 2wh = 12 $ to sub in l = 2w and get $\displaystyle 12 = 4w^{2} + 4wh + 2wh $ and I divided by 2 and solving for h I got $\displaystyle 6 -2w^{2} = (3w)h $ and then $\displaystyle \frac{6-2w^{2}}{3w} = h $ ... so putting this back into the obj. function $\displaystyle V (w) = 2w^{2}(\frac{6-2w}{3w}) $ ... and after taking the derivative and setting it to zero I found that the critical width is $\displaystyle \frac{3}{2} $ , so basically to find the length and height I just plug that back into the constraint? length would equal $\displaystyle 2 * \frac{3}{2} $ and height $\displaystyle \frac{6-2(\frac{3}{2}^{2})}{3(\frac{3}{2})} = h $

For the second problem, is the objective function $\displaystyle A = \frac{a^{2}}{4{\pi}} + \frac{a^{2}-6a+9}{16} $ ?

Re: 2 Optimization Problems

1.) You have correctly found:

$\displaystyle h=\frac{2(3-w^2)}{3w}$

and correctly obtained (after you simplify and factor):

$\displaystyle V(w)=\frac{4}{3}(3w-w^3)$

Your critical value is incorrect though.

2.) Yes, your objective function is correct.