2 w + L = 60 so L =60-2w
A = w L
A = w (60-2w)
A = -2w^2+60 w
2 w^2 -60 w = -A
w^2 - 30 w = -A/2
w^2 - 30 w + 225 = -A/2 + 450/2
(w-15)^2 =-(1/2)(A-450)
w = 15
L = 60-30 = 30
A = 450
a farmer wants to enclose a rectangle garden using the sides of her barn as one side of the rectangle. what is the maximum area which she can enclose with 60ft of fence? what should the dimensions of the garden be to give this area?
Max area which she can enclose with 60ft of fence in sq.ft?
dimensions to give area is 30ft by___ft?
5 answers
If you know any calculus, you can do this much faster
A = -2w^2+60 w
dA/dw = 0 at extreme = -4 w + 60
so w = 60/4 = 15 at max
A = -2w^2+60 w
dA/dw = 0 at extreme = -4 w + 60
so w = 60/4 = 15 at max
I can verify that answer as correct
A batter hits a baseball. The ball's height (in feet) h(t) after t seconds is given by h(t)=-16t^2+160+5
A batter hits a baseball. The ball's height (in feet) h(t) after t seconds is given by h(t)=-16t^2+160+5