Finding a "maximum" area that can be inclosed question.?

Question

If the question was find the mimimum I know I should just use -b/2a, but what should I do when solving for maximum?

A field that borders a river is to be fenced. A rectangular enclosure will be set up with one of its sides being the river. What is the maximum are that can be enclosed?

answer is 31,250 m^2 ?

500 meter of fencing is used for the three sides.

Answer 1

Let l be the length and w be the width.

A = l*w

F = l + 2w since I'm assuming one of the lengths is the river side

F is the amount of fencing which you did not mention in the problem. Anyways, once you plug in F, you can solve for l to get l = F - 2w

Substitute this l into A to get A = (F - 2w) * w

Distribute to get A = Fw - 2w^2

This is a quadratic of the variable w where a = -2 and b = F

w=-b / 2a = - F / 2(-2) = F/4

Now plug this w into l = F - 2w and then multiply l * w to get the maximum area.

Answer 2

You do not give us the length of fence that you have, so we have to stick with proportions.

the depth of the field (from the river edge) is "a" and the width (along the river) is "b".

Fence Length = L = 2a+b (the second b being the river)
Field Area = A = ab

Once you set a fence length, L is a constant, not a variable.

Let us express b in terms of a:
b = L - 2a (from the Length equation)

Substituting b in the second equation (Area) we have
A = a(L - 2a) = L a - 2 a^2

(L is treated as a fixed number)

At maximum (and minimum) the slope of the graph would be zero. We can find the slope with a differential:

dA/da = L - 4 a

Set dA/da to zero:

0 = L - 4a
therefore a = L/4

If the fence length is given, you can now find a. With a and the Length, you can find b. With a and b, you can find area.