Optimization help!

• Nov 1st 2007, 09:30 PM
calculuslyfrustrated
Optimization help!
I can't figure these two problems out for the life of me...this is my last resort lol.

A parcel delivery service will deliver a package only if the length plus the girth (distance around) does not exceed 112 inches. Find the maximum volume of a rectangular box with square ends that satisfies the delivery company's requirements.

and

A fence is to be built to enclose a rectangular area of 300 square feet. The fence along three sides is to be made of material that costs 3 dollars per foot, and the material for the fourth side costs 15 dollars per foot. Find the length L and width W (with W<=L) of the enclosure that is most economical to construct.

Any help would be GREATLY appreciated.

Thanks,
Will
• Nov 1st 2007, 10:20 PM
calculuslyfrustrated
Can anyone at least point me in the right direction? It's due in about 40 minutes and i've been searching and asking to no avail.

• Nov 1st 2007, 11:18 PM
angel.white
Quote:

Originally Posted by calculuslyfrustrated
I can't figure these two problems out for the life of me...this is my last resort lol.

A parcel delivery service will deliver a package only if the length plus the girth (distance around) does not exceed 112 inches. Find the maximum volume of a rectangular box with square ends that satisfies the delivery company's requirements.

and

A fence is to be built to enclose a rectangular area of 300 square feet. The fence along three sides is to be made of material that costs 3 dollars per foot, and the material for the fourth side costs 15 dollars per foot. Find the length and width (with ) of the enclosure that is most economical to construct.

Any help would be GREATLY appreciated.

Thanks,
Will

Okay, I attached a picture of what the box looks like. Since the ends are a square, you know all 4 sides are the same, in this case marked w for width. So the girth is the distance around, meaning the perimeter, or 4w. We are told that the girth (4w) plus the length (l) cannot exceed 112.

So we use the equation
$\displaystyle 4w+l=112$

And we are told to find the maximum volume. Well since volume is the length of all the sides multiplied by eachother, it will be the w*w*l so
$\displaystyle v=w^{2}*l$

Now don't need all these variables, so lets solve for one variable, lets solve for l, and substitute, to cut down on our variables.
$\displaystyle 4w+l=112$
$\displaystyle l=112-4w$

Now we can substitute it into our volume equation:
$\displaystyle v=w^{2}*l$
$\displaystyle v=w^{2}*(112-4w)$

So our volume equation is now solved in terms of w. Now we need to find our endpoints, obviously the width must be greater than zero, because you can't have a box that is less than or equal to zero inches wide so we know that is one boundary, and we can see that if we plug zero in for the length, then 4w=112, so w=28. But since the length cannot be zero either, we know that the width is less than 28. So our domain for w is 0<w<28

If we look at the values of our two endpoints, we can see that
$\displaystyle v(w)=w^{2}*(112-4w)$
$\displaystyle v(0)=0^{2}*(112-4(0))$
$\displaystyle v(0)=0*(112)$
$\displaystyle v(0)=0$

and
$\displaystyle v(w)=w^{2}*(112-4w)$
$\displaystyle v(28)=28^{2}*(112-4(28))$
$\displaystyle v(28)=784*(112-112)$
$\displaystyle v(28)=784*0$
$\displaystyle v(28)=0$

So we know our graph starts and ends at zero. This means that it will go up a bit, then at some point (w,v(w)) it will turn around and start coming back down. Every point where it turns around will have a tangent of 0, so we know that there will be some point w, which has a tangent of zero, this will be the absolute maximum point within our line segment. (See Rolle's Theorem for more on this)

So lets take the derivative of our volume
$\displaystyle v(w)=w^{2}*(112-4w)$
$\displaystyle v(w)=112w^{2}-4w^{3})$
$\displaystyle v\prime(w)=224w-12w^{2})$

Now we can see there will be two points where $\displaystyle v\prime$ will be equal to zero, lets use the quadratic equation to find them:
$\displaystyle 0=224w-12w^{2})$
$\displaystyle 0=56w-3w^{2})$
$\displaystyle w=\frac{-b +/- \sqrt{b^{2}-4(a)(c)}}{2a}$
$\displaystyle w=\frac{-56 +/- \sqrt{56^{2}-4(-3)(0)}}{2(-3)}$
$\displaystyle w=\frac{-56 +/- \sqrt{3136-0}}{2(-3)}$
$\displaystyle w=\frac{-56 +/- \sqrt{3136}}{-6}$
$\displaystyle w=\frac{-56 +/- 56}{-6}$
$\displaystyle w=\frac{0}{-6}$, and $\displaystyle w=\frac{-112}{-6}=\frac{56}{3}$

So we know that at w=0 and w=$\displaystyle \frac{56}{3}$ our line will have a tangent of zero, and therefore be a potential maximum value.

We have already tested for w=0 and saw that it caused volume to go to zero, so this means that when w=$\displaystyle \frac{56}{3}$ that our line which depicts volume will be at it's absolute maximum value.

Now to find that volume, just plug it into the equation:
$\displaystyle v(w)=w^{2}*(112-4w)$

$\displaystyle v(\frac{56}{3})=\frac{56}{3}^{2}*(112-4\frac{56}{3})$

$\displaystyle v(\frac{56}{3})=\frac{3136}{9}*(112-\frac{224}{3})$

$\displaystyle v(\frac{56}{3})=\frac{3136}{9}*(\frac{112}{3})$

$\displaystyle v(\frac{56}{3})=\frac{351232}{27}$

So the maximum volume is $\displaystyle \frac{351232}{27}$cubic inches. or approximately 13008.6 cm^3
• Nov 2nd 2007, 12:33 AM
angel.white
The second question will be basically the same thing, we know our area is 300

so A=300=l*w

We know the cost of our fence will be the perimeter, so the two lengths will be $3 apiece, so the length costs$6 total, and one side of the width will be $15 while the other will be$3, so our width will be $18 total. so cost will be$6l+$18w C=6l+18w Let's put it all in terms of just 1 variable$\displaystyle 300=l*w\displaystyle \frac{300}{w}=l$and substitute$\displaystyle C=6l+18w\displaystyle C=6\frac{300}{w}+18w\displaystyle C=\frac{1800}{w}+18w$And we can see that as the width goes to zero, the length will go to infinity (because a fence with a length of zero has no area) and as length goes to zero, width will go to infinity. A fence that is infinitely long will be mighty expensive, so these points are irrelevant for our considerations. Instead, let's take a look at our critical values, which will be when$\displaystyle C\prime$equals either zero or is invalid (1 over zero)$\displaystyle C=\frac{1800}{w}+18w\displaystyle C=1800w^{-1}+18w\displaystyle C\prime=-1800w^{-2}+18\displaystyle C\prime=-\frac{1800}{w^{2}}+18$Now just figure out where the derivative is at zero$\displaystyle 0=-\frac{1800}{w^{2}}+18\displaystyle -18=-\frac{1800}{w^{2}}\displaystyle -18w^{2}=-1800\displaystyle w^{2}=100\displaystyle w=+/-10$Well our width can't be negative, so that leaves us with just +10 as the only point where the change of cost with respect to width is at zero. We know before and after this point, it will start going back up to infinity, and that's too expensive for us, so we'll check out for width=10 First figure out what length equals$\displaystyle 300=l*w\displaystyle 300=l*10\displaystyle 30=l$So when the width=10, the length=30 and the cost is at it's absolute minimum. Let's plug these values in.$\displaystyle C=6l+18w\displaystyle C=6(30)+18(10)\displaystyle C=180+180\displaystyle C=360$So for the second question, when length=10 feet, width=30 feet, the area will be 300 feet, and the cost will be$360 for the fence, which is smallest it can be while maintaining an area of 300 square feet.
• Nov 2nd 2007, 05:27 AM
CaptainBlack
Quote:

Originally Posted by calculuslyfrustrated
I can't figure these two problems out for the life of me...this is my last resort lol.

A parcel delivery service will deliver a package only if the length plus the girth (distance around) does not exceed 112 inches. Find the maximum volume of a rectangular box with square ends that satisfies the delivery company's requirements.

and

A fence is to be built to enclose a rectangular area of 300 square feet. The fence along three sides is to be made of material that costs 3 dollars per foot, and the material for the fourth side costs 15 dollars per foot. Find the length and width (with ) of the enclosure that is most economical to construct.

Any help would be GREATLY appreciated.

Thanks,
Will

I don't know where the images are hosted, but wherever it is does not have
acceptable security certification for the machines I use. So please do not
use this host again.

RonL
• Nov 2nd 2007, 09:25 AM
angel.white
Quote:

Originally Posted by CaptainBlack
I don't know where the images are hosted, but wherever it is does not have
acceptable security certification for the machines I use. So please do not
use this host again.

RonL

Oh, I didn't realize there were images (i declined the certificates). Hope I didn't miss something important by not viewing the images :/
• Nov 2nd 2007, 02:05 PM
CaptainBlack
Quote:

Originally Posted by angel.white
Oh, I didn't realize there were images (i declined the certificates). Hope I didn't miss something important by not viewing the images :/

There arn't now I have replaced them with Large Times Roman Italic charaters

RonL