Question

Quinn is building an enclosed pen in his backyard he wants the perimeter to be no more than 50 feet he also wants the length to be at least 5 feet longer than the width

Answer 1

Answer:

Option B  width = 5 feet and length = 12 feet

The solution in the attached figure

Step-by-step explanation:

The options of the questions are

Which combination of width and length will meet Quinn’s requirements for the pen?

A.  width = 7 feet and length = 20 feet

B.  width = 5 feet and length = 12 feet

C.  width = 15 feet and length = 10 feet

D.  width = 11 feet and length = 15 feet

Let

x -----> the length of the enclosed pen in feet

y-----> the width of the enclosed pen in feet

we know that

The perimeter is equal to

$P=2(x+y)$

In this problem

$2(x+y)\leq 50$

Simplify

$(x+y)\leq 25$ ----> inequality A

$x\geq y+5$ ---> inequality B

using a graphing tool

The solution is the triangular shaded area

see the attached figure N 1

Remember that

The values of x and y cannot be a negative number

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequality

Verify each case

case A) width = 7 feet and length = 20 feet

so

For y=7, x=20

Check inequality A

$(20+7)\leq 25$

$(27)\leq 25$ ----> is not true

therefore

This combination of width and length will not meet Quinn’s requirements for the pen

case B)  width = 5 feet and length = 12 feet

so

For y=5, x=12

Check inequality A

$(12+5)\leq 25$

$(17)\leq 25$ ----> is true

Check inequality B

$12\geq 5+5$

$12\geq 10$ -----> is true

therefore

This combination of width and length will meet Quinn’s requirements for the pen

case C)  width = 15 feet and length = 10 feet

so

For y=15, x=10

Check inequality A

$(10+15)\leq 25$

$(25)\leq 25$ ----> is true

Check inequality B

$10\geq 15+5$

$10\geq 20$ -----> is not true

therefore

This combination of width and length will not meet Quinn’s requirements for the pen

case D)  width = 11 feet and length = 15 feet

so

For y=11, x=15

Check inequality A

$(15+11)\leq 25$

$(26)\leq 25$ ----> is not true

therefore

This combination of width and length will not meet Quinn’s requirements for the pen

Note If the ordered pair is a solution of the system of inequalities, then the ordered pair must lie on the shaded area

see the attached figure N 2 to better understand the problem