Posted by Lisa on Sunday, March 10, 2013 at 9:09pm.
perimeter = 2 W + L = 900
so
L = (900 - 2W)
A = W L
A = W (900 -2W)
A = 900 W - 2 W^2
now you can complete the square for the parabola and find the vertex or you can use calculus
dA/dW = 0 for max = 900 - 4 W
W = 225
900 = 2 W + L
900 = 450 + L
L = 450
glad you learned calculus and did not have to find the vertex of that parabola? :)
Oh, although I can not see your image I see your note on your earlier question answered by Steve.
If only 100 feet of the building is used then
2 W + L + (L-100) = 900
2 W + 2 L - 100 = 900
2 W + 2 L = 1000
then continue as before
THANK YOU SO MUCH! I got 250 for each dimension :)
Related Questions
Maximum Area - We have 900 ft of fencing and we want to construct a back yard. ...
Precalculus - I have a diagram that has 4 rectangular corrals joined together ...
Precalculus - A barn has 150 feet of fencing and there are 3 rectangular corrals...
college algebra word problem - Here's the problem I was asked: You are ...
calculus - Suppose that 404 ft of fencing are used to enclose a corral in the ...
Pre-Algebra-math - Mr.jones bought 20 yards of fencing to make pen for dog pen ...
Precalculus - There are 4 rectangular corrals of identical dimensions along the ...
PROBLEM SOLVING IN MATHEMATICS - The Andersons are buying a new home. They need ...
math - Chris wants to make an enclosed rectangular area for a mulch pile. She ...
College Algebra - The campus of a college has plans to construct a rectangular ...
For Further Reading