Asked by Rae
Bill wants to fence in his rectangular garden so his precious produce will be protected. Three sides of the fence will be constructed with wire at $2 per linear foot and the fourth side will be constructed of wood fencing at a cost of 6$ per linear foot.If he has $400 to spend on the fencing what dimensions would maximize the area he can plant?
i have my cost function as C(x)= $2(2l+w)+6w
wouldn't i set the cost to 400 after simplifying? how do i get the area..i'm very confused. Help please.
i have my cost function as C(x)= $2(2l+w)+6w
wouldn't i set the cost to 400 after simplifying? how do i get the area..i'm very confused. Help please.
Answers
Answered by
Steve
yes. you know that
2(2l+w)+6w = 600
4l+8w = 600
l+2w = 150
so, l = 150-2w
The area is
a = lw = (150-2w)w = 150w - 2w^2
So, find w which makes a the maximum. That will be at the vertex of the parabola.
2(2l+w)+6w = 600
4l+8w = 600
l+2w = 150
so, l = 150-2w
The area is
a = lw = (150-2w)w = 150w - 2w^2
So, find w which makes a the maximum. That will be at the vertex of the parabola.
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