Consider the following problem: a farmer with 750 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? (a) draw several diagrams illustrating the situation, some with shallow, wide pens and some with deep, narrow pens. find the total areas of these configurations. does it appear that there is a maximum area? if so, estimate it. (b) draw a diagram illustrating the general situation. let x denote the length of each of two sides and three dividers. let y denote the length of the other two sides. (c) write an expression for the total area a in terms of both x and y. a = (375− 5 2y)y

Answer and step-by-step explanation:

For any rectangle, the one with the largest area will be the one whose dimensions are as close to a square as possible.

However, the dividers change the process to find this maximum somewhat.

Letting x represent two sides of the rectangle and the 3 parallel dividers, we have 2x+3x = 5x.

Letting y represent the other two sides of the rectangle, we have 2y.

We know that 2y + 5x = 750.

Solving for y, we first subtract 5x from each side:

2y + 5x - 5x = 750 - 5x

2y = -5x + 750

Next we divide both sides by 2:

2y/2 = -5x/2 + 750/2

y = -2.5x + 375

We know that the area of a rectangle is given by

A = lw, where l is the length and w is the width.  In this rectangle, one dimension is x and the other is y, making the area

A = xy

Substituting the expression for y we just found above, we have

A = x(-2.5x+375)

A = -2.5x² + 375x

This is a quadratic equation, with values a = -2.5, b = 375 and c = 0.

To find the maximum, we will find the vertex.  First we find the axis of symmetry, using the equation

x = -b/2a

x = -375/2(-2.5) = -375/-5 = 75

Substituting this back in place of every x in our area equation, we have

A = -2.5x² + 375x

A = -2.5(75)² + 375(75) = -2.5(5625) + 28125 = -14062.5 + 28125 = 14062.5