Question
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?
Answers
Reiny
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.
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.