calc. dimensions, area

Question

A fence is to be built to enclose a rectangular area of 230 square feet. The fence along three sides is to be made of material that costs 3 dollars per foot, and the material for the fourth side costs 16 dollars per foot. Find the dimensions of the enclosure that is most economical to construct.

Answer 1

Area = 230 = Length times width = LW

L=230/W Let the 4th side that costs $16 per foot be W

The other 3 sides cost $3 per foot

Total Cost = C= 3L + 3W + 16W = 3L+19W = 3(230)/W + 19W = 690/W + 19W

Marginal Cost = derivative = C' = -690/W² + 19

Set it equal to zero and solve for W

-690/W² + 19 = 0

multiply by W²

-690 +19W² = 0

19W² = 690

W² = 690/19

W = square root of 690/19 = about 6.26 feet

L=230/W = 230/(sqr690/19) = about 36.74 feet

check the answer: multiply 36.74 x 6.26 = 229.9924 square feet = virtually 230 given as the area

Also since the side that cost more than 5 times as much would be expected to perhaps be in the range of over 5 times shorter than the cheaper sides, which it is

Answer 2

So what do we know. We need a rectangular area of 230 square feet:

So Area = 230 square feet = w * h

Assume that it is the top has a cost of $16 / foot, while the others have cost of $3 / foot. So the cost of the top is 16w dollars, the bottom 3w dollars, 3h dollars for each side. So the total cost is:

So cost = 16w + 3w + 3 * 2 * h = 19w + 6h

From the area equation we can get either w = 230 / h or h = 230 / w, pick one and substitute into the cost equation. I choose width:

cost = 19 * (230 / h) + 6h

Take the derivative with respect to h:

d cost / dh = -19*230/h^2 + 6

Which look really bad, but wait... The cost is minimum (or maximum) when the derivative is zero. So set the above to zero and solve for h. You see it isn't really bad at all.