Question

I have 47 feet of fencing that I would like to put next to my garage to keep the animals out of my garden. I need only 3 sides to have fencing since my garage provides for the other side. I want the shape to be a rectangular garden and the length to be at least 5 more than twice the width. What would the maximum area be?

Answer 1

If the length is x and the width is w, then we have x >= 2w+5, and x+2w = 47
Assuming the smallest length, we have the area a as
a = xw = w(47-2w) = 47w - 2w^2
da/dw = 47-4w
da/dw=0 when w = 47/4
That means x = 47/2
That gives an area of 2209/8 = 276.125 ft^2

But this means that x is not long enough, as x = 2w
So, since we need x >= 2w+5, let's set x = 57/2, making w = 37/4
Now we have an area of 2109/8 = 263.625
Trying to increase the width does not leave enough fence for the length, so this is the maximum area.
Trying to decrease the width to get more length only lowers the area. For example, using
w = 36/4 = 9 gives x=29, and an area of only 261