SOLUTION: I am having a hard time understanding how to solve this problem. John has 300 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I am having a hard time understanding how to solve this problem. John has 300 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2       Log On


   

Question 51189: I am having a hard time understanding how to solve this problem.
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 by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
2L+2W=300 OR 2(L+W)=300 OR (L+W)=300/2 OR L+W=150 THUS TO MAXIMIZE THE AREA L SHOULD = W SO L&W ARE 75 FEET.
THE MAXIMUM AREA WITH THE MINIMUM PERIMETER IS A SQUARE (ALL SIDES ARE EQUAL).