so you have one length and two widths
width -- x
length --- y
but y + 2x = 212
y = 212-2x
area = xy = x(212-2x) = -2x^2 + 212x
complete the square
area = -2(x^2 - 106x + 2809-2809)
= -2(x-53)^2 + 5618
the largest area is 5618 ft^2 when x = 53 and y= 106
check: 53x106 = 5618
one x lower: 52 x 108 =5616 , smaller
one x higher : 54 x 104 = 5616 smaller
my answer is correct.
A developer wants to enclose a rectangular grassy lot that borders a city street for parking. If the developer has 212 feet of fencing and does not fence the side along the street, what is the largest area that can be enclosed?
1 answer