Problem statement: A farmer has 200 feet of fencing with which to fence off a re

Question

Problem statement: A farmer has 200 feet of fencing with which to fence off a rectangular field. Let the width be x and the length be y. Explain why the area is then A = x*y. Explain why the relationship 2x+2y=200 expresses using all the fencing to make the field. This is called the "constraint" for the problem. Also explain why x and y each have to be at least 0 but at most 100. What would happen if for example x > 100? Simplify the constraint and solve for y, and plug into the A formula, at this step getting a formula for A only using the letter x. The farmer wants to choose x and y so as to maximize the area A. Two methods for this Use calculus on the formula for A found above, so you can say for which x the A is increasing, and for which x A is decreasing, and so get the value of x at which A has its maximum value. Use your graphing calculator to graph the function A of x in the rectangle for x from 0 to 100 and an appropriate range for A. Use the calculator's maximizing command to locate the maximum value of the curve. For this part B of the problem, just say what the screen looks like, and what was reported for the max value.

Explanation / Answer

Area for the rectangle = length * width

Here, length= y

width = x

Therefore, Area should be, A=y*x = x*y

The total fencing used will be equal to the length of perimeter of rectangle which is P = 2( length + width)

Therefore, total fencing used = 2y + 2x = 200

Now, to cover some area , A=x*y

x and y has to be greater than 0.

Total fencing used = 2x+2y = 200

therefore, x+y = 100

hence, if either of x or y greater than 100, it means other one will have to be negative. But dimension of the field can not be negative.

Area, A = x*(100-x) = 100x-x^2

Using calculus, Area will be max. when (dA/dx = 0 )

So, dA/dx = 100-2x = 0

x=50.

Therefore for field to cover maximum area, Dimensions should be width, x= 50ft

and length, y=100-x=100-50=50 ft

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