Question

Consider the following problem: A farmer with 890 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens

Answer 1

Answer:

Total area: 19802.5 ft2

Step-by-step explanation:

The total fencing being used will be the perimeter of the rectangular area plus 3 sides that will divide the total area in 4 pens. These sides will be parallel to the width of the total area, as the width will be the smaller side of the rectangle.

Calling the length L and the width W, we have that:

2L + 2W + 3W = 890

2L + 5W = 890

The total area will be:

A = L*W

From the first equation, we have:

L = 445 - 2.5W

Using this in the area equation, we have:

A = (445 - 2.5W)*W = 445W - 2.5W^2

To find the width that gives the maximum area, we find the vertix of the quadratic equation:

W_vertix = -b/2a

Where a and b are coefficients of the quadratic equation (in our case, a = -2.5 and b = 445). So:

W_vertix = -445/(-5) = 89

The maximum area will be:

A = (445 - 2.5W)*W = (445 - 222.5)*89 = 19802.5 ft2

Answer 2

Answer:

19802.5 ft²

Step-by-step explanation:

Perimeter given is 890ft. According to the figure(see attachment) ,

the equation will be

2x + 5y = 890

5y = 890 - 2x

y= 178- (2/5)x  -->eq(1)

As we know that Area 'A' = xy

Substituting the value of 'y' in above equation

A= x . [178- (2/5)x]

A= 178x - $\frac{2}{5}$ x²

Next is to derive the equation

A' = - $\frac{4}{5}$x + 178

Equating A' to zero in order to find the critical numbers to get your potential minimums & maximums.

- $\frac{4}{5}$x + 178 =0

- $\frac{4}{5}$x  = -178

x = 178/ 0.8

x= 222.5

plugging the above value in eq(1)

(1)=>y= 178- (2/5)(222.5)

y= 178- 89

y= 89

Therefore,

A= x.y => (222.5) (89)  => 19802.5 ft²

The largest possible total area of the four pens is  19802.5 ft²