maximum using completing the square

Question

Is it just me, or this problem does sound weird?

The Parks Department is fencing a rectangular dog-run (a place for dogs to exercise) in a local park. It will be 7 yards longer than 5 times its width. If the Parks Department has allocated 98 yards of prefabricated fence to this project, what are the dimensions of the maximum possible area for the dog-run? You must use completing the square to answer this question.

Attempt: I get too many equations. If the length is 7 + 5 times the width, then we have the following: $$l=7+5w$$ and the perimeter is $$ 98=2w+14+10w=12w+14$$ and we get the width. Where do I use completing the square? If I do not use the information about the length being 7 +5 times the width, then I get $$A=lw$$ and $$98=2l+2w$$ these equations will not give me a concave down quadratic, and hence no use calculating its vertex.

Thanks!

Answer 1

I understand your confusion, since, the constraints. To maximize area given that length is to be 7 more than 5 times the width, the length and width are determinined by using as much of the $98$ yards of fencing:

We have $\mathcal l = 7 + 5w\tag{1}$

So perimeter $98 = 2\mathcal l + 2w$ using $(1)$ gives us $$2(7+5w) + 2w \implies 14 + 12 w = 98 \iff 12 w = 84 \iff w = 7$$

Then expanding on $A = \mathcal l\cdot w = (7 + 5w)w = 7w + 5w^2$

we have $$A = 7w + 5w^2 = 49 + 5\cdot 49 = 6\cdot 49 = 294\;\;\text{sq. yards}$$

Answer 2

Solve $98=2l=2w$ for $l$ in terms of $w$ (or for $w$ in terms of $l$), then substitute into $A=lw$ to get the area as a function of a single variable. However, you'll find that the dimensions maximizing the area through this approach will be those of a square. This is, indeed, a bad problem. Where did you encounter it?