How do you maximize the area, without the use of derivatives, if the length is greater than the one given.

Question

After having a pool built in their backyard, the Harrisons have contracted to have a fence installed around the pool, as required by city ordinance. They want to use 300 feet of fencing and they plan to attach the fence to the house. The back of the house is 90 feet long but they would like to use whatever portion of the house will maximize the area of the rectangular enclosure. The fencing company has quoted a price of $24 per linear foot of fence. This price includes materials and labor for installation. If more than 90 feet are used, what is the the area the Harrisons will have fenced in, the dimensions of the fenced area, the portion of the house that will be used as a part of the enclosure, and the cost? What if less than 90 feet are used?

Answer 1

First thing you should do is draw the problem. Assuming the shape of the barrier will always be a rectangle, you will be dealing with the equation A=l*w no matter how much of the house you use. Lets assume that more than 90ft are used. Since the house is contributing the entire wall, the total barrier is made of 390ft. Thus P=390, and P=2L+2W. This implies 390=2L+2W. We can solve for W here in terms of L => W = 195-L. Then A = L(195-L) = 195L-L². This equation represents an upside down parabola. To find the maximum then all you need to do is find the vertex. This can be done several ways without calculus. One such way is to use the formula x=-b/2a to get the x-coordinate of the vertex => x=-195/-2 = 97.5. This x-coordinate represents the length that gives maximum area. To find the maximum area, plug in 97.5 into the area equation:

A = 195(97.5)-(97.5)²=9506.25 ft²

And width: W = 195-97.5 = 97.5 ft = L

Next, suppose less than 90ft is used. The amount of wall used will be equal to the length of the barrier, L which is between 0 and 90. We can put the whole fencing to use with the added length from the side of the house, thus P=300+L and P=2L+2W => 300+L = 2W+2L. Solving for L this time gives us L = 300-2W. Plugging this into the Area equation gives A = LW = (300-2W)W = 300W-2W². Again this is an upside down parabola and you can take it from here.

Answer 2

Without derivatives, usually the problem results in a quadratic equation (a parabola that points down.) The vertex of that parabola is the maximum.