Pre calc question

Question

We have 800 linear feet of fencing and want to enclose a rectangular field for our dog and then divide it into 4 equal rectangular spaces with internal fences parallel to one of the rectangular sides. What is the maximum area of each space?

Answer 1

Michael,

We can write two equations with the given information. First, area is length times width, so A=Lw.

Secondly, drawing a picture of a field that is L ft. by w ft. will show there are five of one length, which we'll designate w, and two of the other measurement L.

Which leads to our second equation, giving the total perimeter, or length of fencing for the enclosures: 5w+2L=800.

The general strategy with this sort of problem is to solve one equation for one variable and replace that variable in the other equation, so we end up with one equation and one variable:

Solve 5w+2L=800 for L:

Subtract 5w and divide by 2 to get: L=(800-5w)/2, or L=400-(5/2)w. Substitute this for L in the formula for area to get: A= [(400-(5/2)w]w. Distribute the w and put the squared term first to attain: A= -(5/2)w^2 +400w.

Find the vertex of this equation by whichever means you prefer: graphing, converting to vertex form, solving for w with -b/2a, etc.

The vertex is at (80, 16,000). So the max area of the entire enclosure is 16,000 square feet when w=40 ft..

Plug 40 into the linear equation 5w+2L=800 to find that w=40 ft. when L=200 ft.

Since this represents the entire space, we can answer the original question by dividing the total area by 4: 16,000/4=400 square feet per enclosure. They are each 40 by 100 ft.

I hope this helps!

Let me know if you have any questions, and have fun mathing!

LizZ