SOLUTION: 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

Question 71238: 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 formula to find the maximum area.

Found 2 solutions by checkley75, galactus:
Answer by checkley75(3666) (Show Source): You can put this solution on YOUR website!
THE MAXIMUM AREA OF ANY RECTANGLE IS A SQUARE. THUS:
2X+2X=300
4X=300
X=300/4
X=75 FEET PER SIDE.
THIS ASSUMES THE PATIO IS A STAND-ALONE STRUCTURE.
IF IT IS ATTACHED TO A HOUSE THEN THERE ARE ONLY THREE SIDES THAT NEEDS TO BE ENCLOSED. THUS THE SOLUTION HERE IS
3X=300
X=300/3
X=100 FEET ON A SIDE.
Answer by galactus(183) (Show Source): You can put this solution on YOUR website!
Let x = the width and y = length of the patio. The perimeter of the patio can be represented by . The area by . See so far?.
Solve 2x+2y=300 for y and sub into the area formula.

Now, sub into area, xy:

You probably know the x-coordiante of the vertex is given by
a=-1 and b=150

Therefore, the width x, is 75.
Subbing gives us y=75 also.
Therefore, max area is achieved when the width is 75 and the length is 75.
It can be shown through calculus that the max area is achieved when the area is a square. Which you have. 75X75 square patio.
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