SOLUTION: 4) 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 o

Question 53231: 4) 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. Use the vertex form to find the maximum area.
Answer:
Show work in this space.
Thank You
Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
P=2L+2W
Let the length =L
We only want one variable, so plug 300 in for p and solve for W (your width).
300=2L+2W
300-2L=-2L+2L+2W
300-2L=2W
(300-2L)/2=2W/2
300/2-2L/2=W
150-L=W
A=LW We only want one variable, so substitute the value above in for W.
A=L(150-L)
A=150L-L^2
A=-L^2+150L
A=-(L^2-150L)
A=-(L^2-150L+(150/2)^2)+(150/2)^2
A=-(L^2-150L+75^2)+75^2
A=-(L-75)^2+5625
The vertex of a parabola represents the maximum or minimum of the parabola.
The vertex form of a parabola is f(x)=a(x-h)^2+k where the vertex is (h,k).
In our case the vertex is (75,5625)
Our maximum L value we get from the h coordinate of our vertex=75 ft. That's our maximum length.
Our maximum W (width) value is 150-L=150-75=75 ft.
Our maximum area we get from the k value of our vertex= 5625 ft.
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