The pig pen: Suppose that a rectangle pig pen is 4 feet longer than it is wide and its area is more than 32 square feet. The question is the dimension of the pen.

Question

Let 𝑥 be the width or the pen.

1. Using 𝑥, give the pig pen length and area.

2. Explain why 𝑥 2 + 4𝑥 − 32 > 0

3. Sketch the graph of the function 𝑦 = 𝑥 2 + 4𝑥 − 32 and shade the 𝑥 −values that satisfy the inequality : 𝑥 2 + 4𝑥 − 32 > 0.

What could the dimensions be ?

The farmer gets more pigs and increases the dimensions of his pen. The pig pen is now 10 feet longer than it is wide and its area is exactly 96 square feet

4. Explain why 𝑥 2 + 10𝑥 − 96 = 0

5. Solve this equation

6. What are the new dimensions?

Answer 1

length = 4 + x

(4+x)x > 32

x^2 + 4x > 32

-x^2 + 4x -32 > 0

(x+8)(x-4) > 0

either both factors are positive or both are negative. They can't be opposite signs so

x+8>0 and x-4>0 or x+8<0 and x-4<0

x>-8 and x>4 or x<-8 and x<4

x>4 or x<-8

ignoring a negative width, the width must be > 4

Length > 8

dimensions are 4 by 8 or greater

if length = width + 10 = x+10 then area = length times width = (x+10)x = x^2 +10x = 96

or

x^2 +10x -96=0

(x-6)(x+16) = 0

x=6

length =6+10 = 16

new dimensions are 6 by 10