S Lesson 19 Modeling with Fur x c Chegg study iouised Sciuto

S Lesson 19 Modeling with Fur x c Chegg study iouised Sciuto x 15673503 A graphing calculator is recommended. Consider the following problemt A has 360o n of fencing and wants to fence om a rectangular field that borders a straight river. He does not need a fence along the nver (see the figure) What are the mensions of the field largest area that he can rence (Let x be the width of the read in feet and Ibe the length of the neld in feet) di the problem by drawing several diagrams ilustrating the situation calculate the area of each oonnguration, and use your results to estimate the dumersians Experiment the largest possible field. (Round your answers to the nearest hundred feet.) (b rind a runotion that models the area of the feldin terms of ene of ts des (c) use your mode to solve the problem. and oompare wth your answer Part

Solution

(b) Let the width of the field be x ft. Then the length of the field is 3600-2x ft so that the area of the field is A(x) = (3600-2x)*x = 3600x-2x².

(c) We know that, for A(x) to be maximum, dA/dx = 0 and d²A/dx² should be negative. Here, dA/dx = 3600 -4x and d²A/dx² =-4. Thus, if dA/dx = 0, then 3600-4x = 0 so that x = 3600/4 = 900. Further, regardless of value of x, d²A/dx² is negative. Hence, A(x), the area of the field, will be maximum if the width of the field is 900 ft. The length of the field will be 3600-2*900 = 3600-1800 = 1800ft. Then, the area of the field = 900*1800 = 1620,000 sq.ft.

(a).

1. When the width = 750 ft, then the length of the field is 3600-2*750 = 3600-1500 =2100 ft and the area of the field = 750*2100 = 1575,000 sq.ft.

2. When the width = 800 ft, then the length of the field is 3600-2*800 = 3600-1600 =2000 ft and the area of the field = 800*2000 = 1600,000 sq.ft.

3. When the width = 850 ft, then the length of the field is 3600-2*850 = 3600-1700 =1900 ft and the area of the field = 850*1900 = 1615,000 sq.ft.

4. When the width = 900 ft, then the length of the field is 3600-2*900 = 3600-1800 =1800 ft and the area of the field = 900*1800 = 1620,000 sq.ft.

5. When the width = 950 ft, then the length of the field is 3600-2*950 = 3600-1900 =1700 ft and the area of the field = 950*1700 = 1615,000 sq.ft.

Thus, for the largest area of the field, x = width of the field = 900 ft. and the length of the field is 1800 ft.