Mathematics

I am studying for my exams and do not know if I tackled correctly this word problem. Would appreciate any help. Thank you

"Tony and Doris wish to fence off part of their garden to create a
rectangular enclosure for their dogs. One side of the
enclosure will be provided by part of an existing wall. They have 220m of
fencing material available and wish to enclose as large an area as possible.
The length and width of the enclosure are denoted by l and w.

Question a): Write down an expression for the area A in terms of w and l. Also, write down an expression relating the length of the fencing to w and l.

Answer a): Area (A) = wl / Length of fencing (F) = 2w + l

Question b): Use the result from part (a) to write A in terms of w only.

Answer b):
F = 2w + l
220 = 2w + l
l = 220 - 2w
So A in terms of w is
A = wl
A = w(220 - 2w)
A = 220w - 2w^2

Question c): Use differentiation to determine the value of w that gives the maximum value for the area A enclosed and show that this is a local maximum. What is the maximum rectangular area that Tom and Barbara can enclose?

Answer c):
A = 220w - 2w^2
dA/dw = 220 - 4w

At a stationary point, we have
dA/dw = 0 = 220 - 4w
Solving for w gives, w = 55.
So the maximum area A achieved is when w = 55m.
Therefore, substituting w = 55 into the formula gives
F = 2w + l
220 = 2(55) + l
l = 110.
Hence the maximum rectangular area they can enclose is 55 * 110 = 6050m^2.

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  1. you are correct.

    Just FYI to save time, maximum area is achieved when the fencing is divided equally among lengths and widths. So, 110 m are allocated to each group.
    2 widths means w = 55
    1 length means l = 110

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