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

Algebra ->  Quadratic Equations and Parabolas -> 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       Log On


   

Question 52433: 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.
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
THE MAXIMUM AREA FOR A RECTANGLE IS A SPECIAL RECTANGLE CALLED A SQUARE. THUS
300=X^2 OR X=SQRT300 OR X=17.320508 FEET
PROOF 17.320508*17.320508=300 OR 300=300