The fencing-length information
gives me perimeter. If the length of the enclosed area is L
and the width is w,
then the perimeter is 2L
+ 2w = 500,
so L
= 250 – w. By
solving the perimeter equation for one of the variables, I can substitute
into the area formula and get an equation with only one variable:A =
Lw = (250 – w)w = 250w – w2
= –w2 + 250wTo find the maximum, I
have to find the vertex (h,
k).h =
–b/2a
= –(250)/2(–1) = –250/–2 = 125In my area equation, I
plug in "width" values and get out "area" values.
So the h-value
in the vertex is the maximizing width, and the k-value
will be the maximal area:k =
A(125) = –(125)2 + 250(125) = –15 625 + 31 250 =
15 625The problem didn't ask
me "what is the value of the variable w?",
but "what are the dimensions?" I have w
= 125. Then the length
is L
= 250 – w = 250 – 125 = 125.
