Asked by Jessica
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
Answered by
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.