A nursery school wants to fence in a rectangular playground. one side of the playground is along a busy street, and needs taller fencing that costs $10 per foot. The fencing for the other three sides costs $7 per foot. If the school has a budget of $6000, what are the dimensions of the playground of largest area that they can build?

Could I just make one length x and the width y (or vice versa), solve for y, and use f(x) = x * y for the area, or would there be a problem since one side costs more than the others?

Question

A nursery school wants to fence in a rectangular playground. one side of the playground is along a busy street, and needs taller fencing that costs $10 per foot. The fencing for the other three sides costs $7 per foot. If the school has a budget of $6000, what are the dimensions of the playground of largest area that they can build?

Could I just make one length x and the width y (or vice versa), solve for y, and use f(x) = x * y for the area, or would there be a problem since one side costs more than the others?

anonymous · Answer

it would be a problem because one side costs more, and you're limited by a $6000 budget :(

anonymous · Answer

So how would I solve it? Please help :(

anonymous · Answer

What class is this from?

anonymous · Answer

as in Adv Algebra, pre-cal, calculus?

anonymous · Answer

Calculus

anonymous · Answer

Than this is an Optimization problem. Here's how I would set it up:

6000 = 7L + 10L + 7W + 7W

6000 = 17L + 14W

6000 - 14W = 17L

(6000 - 14W) / 17 = L

A = (L)(W)

See where you can go from there ;)

anonymous · Answer

Great! Thank you!