SOLUTION: Could someone please help me with this Thanks!!!!!!! ) John has 300 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times wi

Question 42205This question is from textbook college algebra gary rockswold
: Could someone please help me with this Thanks!!!!!!!

) John has 300 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 times width). He wants to maximize the area of his patio (area of a rectangle is length times width). What should the dimensions of the patio be, and show how the maximum area of the patio is calculated from the algebraic equation.
Show clearly the algebraic steps which prove your dimensions are the maximum area which can be obtained.
Answer:
Show work in this space. This question is from textbook college algebra gary rockswold

Answer by psbhowmick(878) (Show Source): You can put this solution on YOUR website!
Let length = L ft, width = W ft.
Then perimeter = 300 = 2(L + W)
or L + W =150 _____(1)

Now, area is given by
A = W x L
= (150 - L) x L
=
=
or ________(2)

For 'A' to be maximum, has to be minimum.
But it is a real square and hence cannot be negative.
So its minimum value is zero.
So,
or L = 75

We have obtained L = 75. Putting this value in (1) we find W = 75.

Putting L = 75 in (2) we will get the maximum value of 'A'.
A = 5625

So the maximum area of the patio is 5625 sq ft and its dimensions are 75 ft x 75 ft. In other words for maximum area, the patio is a square with side 75 ft.
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