Question

A large lot in a park is going to be split into two softball fields, and each field will be enclosed with a fence.

• A rectangular lot, wider in width than it is tall in height, is shown. A vertical line begins about halfway across the top of the rectangle extends down to bottom side, splitting the lot in two.
• The top side length is labeled y.
• The left and right side lengths are labeled x, as well as the middle vertical line.

The parks and recreation department has 1,800 feet of fencing to enclose the fields. What dimensions will enclose the greatest area?

(a)

Fill in the following equation.

· x +

· y = 1,800

(b)

Solve the equation in part (a) for y.

y =  +

(c) · x

Using

A = x · y,

write an equation for the area A in terms of x.A =

(d)

What values of x and y will maximize the area of the enclosures?

x=y=

(e)

What is the maximum area (in square feet)?

sq ft

Answer 1

Salon here we conside? Ilu Sketch. In görninni id) sönce, neve secrcation department aus 100 of pencing lupp curlose the field wind we lemow Hiat les total fencing sequired should be equal to the perimeter of Hii Brildi.10, ormulating in above stotomond as muthimalica/strotomcut, we have - Perimeter of AEND + Perimeter of (EBCA) - EF 100 ie (2x+42 + (2X+Y) -- M21901) z 3x+2y = 1900 ت ن ن ن ن ن ن ن HV b) sowing equation 0 , for y, we have 2y = 1020-34 | 1/9 = ( 100 - 301 c) sinu the total area of the enclosed field Azxy - subtituting the value of y from ② in the above egn, we get Aza(+ 11800-3x) = quox – 32 x? 7 Az 9000-3/2012 UUUUUUUUUUUUUUUUUUUU

c d ). Now, we have problem man . A = 9004 - 8/2 x 2 st. 20. For maxima o minima of A, A! 20 ! ! ! ! ! 2 900 - 32 20 - 2 300), 2-3<0 & The point xzz00 is the point 1430 of maxima. Thus, by equation (2), y z ? (1800-32) = Ž ( 1020 - 3.300) 2450. Thus, the area will be max. for a=300, y = 150. se). The maximum area will be given by Amaro z 300 x 450 = 13 5000 sq. feet a l Amax? 135000 sq. fect 222.2220 2.2.A....