SOLUTION: Suppose you have 80 ft of fence to enclose a rectangular garden.The function A=40x-x^2 gives you the area of the gardne in square feet where x is the width in feet.
a. What width,
Question 419440: Suppose you have 80 ft of fence to enclose a rectangular garden.The function A=40x-x^2 gives you the area of the gardne in square feet where x is the width in feet.
a. What width,x, gives you the maximum garden area?
b. what is the maximum area? Found 2 solutions by Gogonati, lwsshak3:Answer by Gogonati(855) (Show Source):
You can put this solution on YOUR website! Since the width of rectangular is x ft the length of this rectangle is ( 80-2x)/2
and the enclosed area is x(80-2x)/2=40x-x^2. A=40x-x^2 is an downward parabola and the value y of its vertex is the maximum area.
A=-x^2+40x the coordinate x=-b/2a=-40/-2=20 and the A=-20^2+40(20)=400.
Answer: The maximum enclosed area is 400 sft.
You can put this solution on YOUR website! Suppose you have 80 ft of fence to enclose a rectangular garden.The function A=40x-x^2 gives you the area of the gardne in square feet where x is the width in feet.
a. What width,x, gives you the maximum garden area?
b. what is the maximum area?
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Standard form for parabola:y=(x-h)^2+k,(h,k) being the (x,y) coordinates of the vertex, from which you can determine the maximum or minimum and the x-coordinate where it occurs.
A=40x-x^2
A=-(x^2-40x)
complete the square
A=-(x^2-40x+400)+400
A=-(x-20)^2+400
ans:
The width which gives the maximum area=20 ft
The maximum area = 400 sq ft
With dimensions 20 ft by 20 ft=80 ft of fencing
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