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Jim has 120 m of fencing that he wants to use to enclose a rectangular field beside a river. If no fencing is needed along the river, what is the maximum area of the field that Jim can enclose?

User Arkonautom
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Explanation:

x = length

y = width

let's assume that a length is beside the river (but we will see that it does not matter).

for 2 variables we need to equations to solve.

one is about the perimeter. we know the perimeter is 120 m.

120 = 2×y + 1×x = x + 2y

therefore,

x = 120 - 2y

and then we have the area we want to maximize :

area = x × y

we use the found identity of the first equation in the second :

area = f(y) = (120 - 2y) × y = 120y - 2y²

the zeros of the first derivative of a function indicate the extreme points of the function.

the sign of the second derivative of that function at these points tells us, if it is a local max. or a min. if the second derivative is positive at the point, then it is a min.

if it is negative, it is a max.

f'(y) = 120 - 4y = 0

4y = 120

y = 120/4 = 30 m

x = 120 - 2y = 120 - 2×30 = 60 m

f''(y) = -4

so,

f''(30) = -4 and therefore negative (it is a max.).

this maximum area is therefore

30×60 = 1800 m²

and so, we see, if we make a width to be a side on the river, we only need to exchange x by y and y by x in all the equations. and we get the same equations with the same numbers and solutions.

it tells us that the side along the river is 60 m long. and the other side type is 30 m long.

User Nightloewe
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