Q8 A developer wants to enclose a rectangular grassy lot tha

Q8. A developer wants to enclose a rectangular grassy lot that borders a city street for parking. If the developer has 320 feet of fencing and does not fence the side along the street, what is the largest area that can be enclosed?

   a. 25,600 ft²
   b. 19,200 ft²
   c. 12,800 ft²
   d. 6400 ft²

Solution

7. A slant (oblique) asymptote occurs when the polynomial in the numerator is a higher degree than the polynomial in the denominator. Here, both the numerator and the denominator of h(x) are of the same degree (2) so that there is no oblique asymptote. The answer d is correct.

8. Let the length and the width of the rectangular plot be x and y feet respectively. Then the area (A) of the plot is x y and the perimeter is 2x + 2y. Since the side along the street is not to be fenced, and since there is only 320 feet of fencing, we have x + 2y = 320 so that y = ½(320 – x). Then the area A = ½ x(320 –x) = -1/2 x² + 160x. If A is to be maximum, then dA/dx = 0 and d²A/dx² should be negative. Here, dA/dx = -x + 160. Thus, if dA/dx = 0, then –x + 160 = 0 or, x = 160. Also, d²A/dx²= -1 which is negative. Thus, A is maximum when x = 160. Then y = ½( 320 –x) = ½(320-160) = 160/2 = 80. Thus, the maximum area of the rectangular grassy lot that can be enclosed on 3 sides ( other than that along the street ) is A = x y = 160*80 = 12800 sq feet. or 12800 ft². The answer c is correct.