Question

A farmer is building a fence to enclose a rectangular area against an existing wall, shown in the figure below. A rectangle labeled, Fenced in Region, is adjacent to a rectangle representing a wall. A rectangle labeled, Fenced in Region, is adjacent to a rectangle representing a wall. Three of the sides will require fencing and the fourth wall already exists. If the farmer has 184 feet of fencing, what are the dimensions of the region with the largest area?

Answer 1

Answer:

  46 ft by 92 ft

Step-by-step explanation:

The largest area is enclosed when half the fence is used parallel to the wall and the other half is used for the two ends of the fenced area perpendicular to the wall. Half the fence is 184 ft/2 = 92 ft. Half that is used for each end of the enclosure. So, the dimensions are 46 ft by 92 ft.

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Working

Let x represent the dimension parallel to the wall. Then the other dimension is ...

  (184 -x)/2

and the total area is ...

  Area = x(184 -x)/2

This is the equation of a parabola that opens downward. It has zeros (x-intercepts) at x=0 and x=184. The vertex (maximum area) is located on the line of symmetry, halfway between those x-intercepts. The maximum area is found where x=184/2 = 92.

Then the other dimension is (184-92)/2 = 46.

Answer 2

The dimensions of the region with the largest area are 92 ft by 42 ft.

 Let x = length of side parallel to the wall

    y= length of each side perpendicular to the wall

Total fencing area =184 feet

$x+2y=184\\2y=184-x$

$y=$ $\frac{184-x}{2}$

The total area of the rectangular field  

Area =length ×breadth

Area = $x\frac{(184 -x)}{2}$

Differentiate it w.r.t x ,we get

$\frac{dA}{dx} =92-x$ and $\frac{dA}{dx} =-1$

Put the first derivative equals to zero , we get

x=92

$\frac{dA}{dx} =-1<0$ at x=92

So, the area is maximum for x=92 feet

Now, the other dimension is

$y=\frac{184-92}{2} \\=46 feet$

To maximize the area, each side perpendicular to the wall should have length y = 46 ft and the side parallel to the wall should have length x= 92 ft.  

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