Question 527254: a rectangle has a perimeter of 40 meters. find the dimensions of the rectangle with the maximum area.
Please go step-by-step
I started with the equation 2x+2y=40 and made it x+y=20, but I am supposed to use quadratic equations with only one variable... so I made them both x. it became x(-x+20) and I got -x^2+20x. then I solved -b/2a and got 10. but when I plug that in I get y to equal 100, which is not right. please help me understand. thank youu!
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! your sort of on the right track...
P = 40 ___ 2L + 2W = 40 ___ L + W = 20
with one variable; if x is L, then W is 20-x ___ so the area is L*W or 20x-x^2
you used the axis of symmetry to find the max, which is 10
the y that you found (from y = -x^2 + 20x) is the area (which maxes at 100)
L is 10, and so is W (20 - 10)
the rectangle with the max area is a square
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