Help with this optimization calculus problem?

I think the *more* costly side is 16 vice 12.
So, same derivation as above but answer is
W = 10*sqrt(12/11)
L = 20*sqrt (11/12)

In order to answer optimization problems, you need to set-up first the equation for what you need to optimize. In this problem, you need to optimize cost. Thus, if you let

l = length
w = width
C = cost

then

C = 6(l + l + w) + 16w
C = 12l + 22w

Note that in this problem, I assumed that the fourth side, which is more expensive to fence, is a width.

Now, we also know that

A = lw
200 = lw

because the area is 200 sq ft. This information is important, as it will allow us the following transformation,

l = 200/w

which ultimately leads to the substitution

C = 12(200/w) + 22w
C = 2400/w + 22w

Note that we need the cost to be a function affected by only 1 variable, and that is the reason for the substitution. Note that if we did not substitute for length, then we would have 2 changing variables on the right side of the cost equation: the length and the width. You know that 1 changes with the other, because they have to maintain the 200 sq ft of area. Now that we have set-up the equation where we would optimize the cost, and finished with the substitution so that it is only determined by 1 variable, we now begin to differentiate.

C' = -2400/w^2 + 22 {I hope that you have no problem getting the derivative}

Since the optimum value occurs when the first derivative is set to 0, we then have

0 = -2400/w^2 + 22
22w^2 = 2400
w^2 = 1200/11
w = sqrt(1200/11), or
w = 20/11 * sqrt(33) {this was obtained by simplifying the one above, and by rationalizing it}

To determine the length, we have

l = 200/w
l = 200/ [20/11 * sqrt(33)]
l = 110/sqrt(33)
l = 10/3 * sqrt(33)

Thus, the dimensions of your rectangle is

20/11 * sqrt(33) by 10/3 * sqrt(33)

Hope this helps!!!!!

very elegant solution