A farmer has 110 metres of fencing to fence off a rectangular area. Part of one side is a wall of length 15m. Find the dimensions of the field that give the maximum area. Answers: length and width = 31,25m

Question

Thank you so much for a huge help.

Answer 1

Since there is additional 15 m which is part of one side of the wall, we can say that the total fencing material is
110 + 15 = 125 m
Recall that perimeter of rectangle is just
P = 2(L + W)
where L is the length and W is width.
Substituting,
125 = 2(L + W)
62.5 = L + W, or
L = 62.5 - W
Now, recall that area of rectangle is
A = L*W
Since we have an expression for length, we can substitute it here,
A = (62.5 - W)(W)
Since we want area to be maximized, we differentiate Area with respect to Width and equate to zero (since at maximum, the slope is zero):
A = 62.5*W - W^2
dA / dt = 62.5 - 2W
0 = 62.5 - 2W
2W = 62.5
W = 31.25 m (width)
Substituting this to get the length,
L = 62.5 - 31.25
L = 31.25 m (length)

Hope this helps~ :3

Answer 2

*lol, sorry it shouldn't be dA/dt, rather
dA/dW = 62.5 - 2W

but it won't affect the answer anyway. (it's just that you might wonder where the variable t came from) ^^;