Question

SCALCET8 4.7.011. Consider the following problem: A farmer with 950 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

Answer 1

Answer:

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

Step-by-step explanation: