Asked by Vin
My friend has $80 to spend on a fence for her rectangular garden. She wants to use cedar fencing which costs $8/foot on one side, and cheaper metal fencing which costs $2/foot for the other three sides. What are the dimensions of the garden with the largest area she can enclose? What is the largest area that can be enclosed?
I have the perimeter as 10x+4y=80. From this I get y=20-(5x/2). So I plug that into the formula WidthxLength for Area and get the equation x(20-5x/2), which gives me 8 and 0 as the values for x, of which 8 is the only feasible. But when plugged back in to the equation for y, I get 0 which is obviously impossible. What a I missing?
I have the perimeter as 10x+4y=80. From this I get y=20-(5x/2). So I plug that into the formula WidthxLength for Area and get the equation x(20-5x/2), which gives me 8 and 0 as the values for x, of which 8 is the only feasible. But when plugged back in to the equation for y, I get 0 which is obviously impossible. What a I missing?
Answers
Answered by
Steve
Your algebra is correct. Your logic is wrong, however.
x=0,8 are roots of the function. That is, where the area is zero.
If you rewrite it as
20x - 5/2 x^2
you see it is a parabola with its vertex at x = -b/2a = 20/5 = 4
So, y=10
and the largest area is 4*10 = 40
http://www.wolframalpha.com/input/?i=x%2820-5x%2F2%29
x=0,8 are roots of the function. That is, where the area is zero.
If you rewrite it as
20x - 5/2 x^2
you see it is a parabola with its vertex at x = -b/2a = 20/5 = 4
So, y=10
and the largest area is 4*10 = 40
http://www.wolframalpha.com/input/?i=x%2820-5x%2F2%29
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