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
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Question 24390: 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?
I have tried over and over again to form some sort of equatin and can't come up with one. What am I doing wrong? Answer by queenofit(120) (Show Source):
You can put this solution on YOUR website! Have you tried this.
Perimeter = 300
width = w
length = l= 2w in a rectangle the length is twice the width
equation;
2(2w) + 2w = 300
4w+2w=300
6w=300
w=50
l=2w=100
the demensions would be 50 feet by 100 feet.
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