2 Optimization Problems

**AZach** · Nov 26th 2012, 07:48 AM

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 $V (l, w, h) = lwh$ and since $l = 2w$ ... $V (w, h) = 2w^{2}h$

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

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

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

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

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

Thanks!

**MarkFL** · Nov 26th 2012, 12:40 PM

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:

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

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 $a$ , 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.

**AZach** · Nov 26th 2012, 05:30 PM

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

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

**MarkFL** · Nov 26th 2012, 05:57 PM

1.) You have correctly found:

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

and correctly obtained (after you simplify and factor):

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

Your critical value is incorrect though.

2.) Yes, your objective function is correct.

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