Three of the sides will require fencing and the fourth wall already exists. If the farmer has 144 feet of fencing what is the largest area the farmer can enclose ?

Complete question is;

A farmer is building a fence to enclose a rectangular area against an existing wall. Three of the sides will require fencing and the fourth wall already exists. If the farmer has 144 feet of fencing, what is the largest area the farmer can enclose?

A_largest = 2592 ft²

Step-by-step explanation:

Let the width which is going to be perpendicular to the already existing fourth wall be denoted by y. Since there will be 2, it means total width = 2y.

Since the farmer has 144 ft of fencing and the total width is 2y,it means that the length of the fence will be expressed as;

Length = 144 - 2y

Area of rectangular fencing is;

A = y(144 - 2y)

A = 144y - 2y²

To find the largest area, we will find the first derivative and the equate to 0 after which we plug the value of y back into the area equation to get it.

dA/dy = 144 - 4y

At dA/dy = 0;

144 - 4y = 0

y = 144/4

y = 36 ft

Thus, largest area is;

A_largest = 144(36) - 2(36)²

A_largest = 2592 ft²

2592

Step-by-step explanation:

If the two equal lengths on the left and right of the region have length x, then the third length on the lower side of the region has length 144−2x (because the total length of the fencing is 144.

Therefore, in terms of the side length x, the area of the region is

x(144−2x)=144x−2x2

This is a quadratic function, so to maximize it we should find the y-value of its vertex. Recall that the x-value of the vertex of the parabola ax2+bx+c is given by the formula −b2a. So in this case, we have a=−2 and b=144, so the x-value of the vertex is

−b2a=−(144)2(−2)=36

Plugging this into the formula for the area, we find that the maximum area is

144(36)−2(36)2=5184−2592=2592

central angle of a triangle has the same measure as its intercepted arc.

step-by-step explanation:

113.42

i hope that is right.

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