Have a tough time solving these set of questions

Question

A contractor has 250 feet of enclosure material (similar to a fence but without holes) is to be installed

around a rectangular surface that will be used for many different sports and activities. The

installation plans have one edge of the playing surface up against an existing wall, so only the other

three sides need the enclosure material installed.

A. What is the width and area of a rectangle surface with 180 ft width and perimeter of 250 ft

A.1 what is the width and area of a rectangle surface with 150 ft width and perimeter of 250 ft

b. If the width of the playing surface is w feet, write a function A(w) which represents the area of

the playing surface as a function of the width.

C. Find the exact width of the playing surface that will maximize the area (do this algebraically).

d. Determine the maximum area that can be obtained with 250 feet of enclosure material.

Answer 1

Hi Lee,

Have you started in on the first few sub-questions yet? Even if you're struggling with the optimization work, you can always try for partial credit by answering some of the math.

For parts A and B, this is pretty simple PreAlgebra :) you know that the formula for area of a rectangle is:

A = L × W, where A is area, L is the given length, and W is the given width.

And you know that to find perimeter, you would use the formula:

P = 2L + 2W, to represent P or perimeter that is made up of 2 of the "length" sides and 2 of the "width" sides.

In this case, it looks like they're telling you, one of the sides is already there, and we don't have to worry about it. It's hard to tell exactly what they're going for, but my guess is that one of the "width" sides is preexisting, so with our 250 ft of material, our new Perimeter formula would be:

P = 2L + W, to represent the two sides sticking out of the wall and the one W width that connects those two sides.

So, if you're given a rectangle with 180 ft width and 250 ft perimeter, you can find the missing quantity L with the Perimeter formula:

P = 2L + W

250 = 2(L) + (180)

250 = 2L + 180

70 = 2L

35 = L

and you would be able to make an enclosure that has 3 pieces, 2 of those "length" pieces attached to the wall that are 35 ft long each, and then at their ends, they're connected by a 3rd "width" piece that's 180 ft long.

The second sub-question, part A1, asks you to do the same thing over again but instead of using 180 as your width, use 150. See how that changes your answer. To check it, I would think to myself, "okay, my long side is going to be a little bit shorter this time. That should give me some extra material to put towards the other two short sides, and so I would expect my new L to be a little bit bigger than the first L that we calculated." Does that match up with the answers you get?

Try to see if you can get that far, and if that helps you move through the rest of the problem. If you get stuck again, or want to go back over anything I've explained here, let me know and I'll be happy to help!