To solve the problem, let's address the constraints systematically.
1. Dimensions of the Car:
- The length [tex]\( l \)[/tex] must be 3 inches greater than the width [tex]\( x \)[/tex]. Thus, we can write:
[tex]\[ l = x + 3 \][/tex]
- The width [tex]\( x \)[/tex] must be at least 2 inches greater than the radius of the wheels [tex]\( y \)[/tex]. Hence:
[tex]\[ x \geq y + 2 \][/tex]
2. Cost Constraints:
- The cost of the base is calculated based on its area. If the length is [tex]\( l \)[/tex] and the width is [tex]\( x \)[/tex], then the area [tex]\( A \)[/tex] is [tex]\( x \times l \)[/tex]. The cost per square inch is [tex]\( \$0.50 \)[/tex], so the total cost for the base is:
[tex]\[ \text{Cost of base} = 0.50 \times x \times l \][/tex]
- The radius of each wheel is [tex]\( y \)[/tex], and each of the 4 wheels costs [tex]\( \$2.25 \)[/tex] per inch of radius. Thus, the cost for the wheels is:
[tex]\[ \text{Cost of wheels} = 4 \times 2.25 \times y = 9y \][/tex]
- Combining these, the total cost of the car must not exceed [tex]\( \$50 \)[/tex]:
[tex]\[ 0.50 \times x \times l + 9y \leq 50 \][/tex]
- Substituting [tex]\( l = x + 3 \)[/tex]:
[tex]\[ 0.50 \times x \times (x + 3) + 9y \leq 50 \][/tex]
[tex]\[ 0.50(x^2 + 3x) + 9y \leq 50 \][/tex]
[tex]\[ 0.50x^2 + 1.5x + 9y \leq 50 \][/tex]
Now we can translate these constraints into a system of inequalities. Given [tex]\( x \geq y + 2 \)[/tex], and combining it with the cost constraint, we have:
[tex]\[ x \geq y + 2 \][/tex]
[tex]\[ 0.50x^2 + 1.5x + 9y \leq 50 \][/tex]
Which matches system D. Therefore, the correct system of inequalities that John and his son should use is:
[tex]\[
\begin{array}{l}
x \geq y + 2 \\
0.50 x^2 + 1.5x + 9 y \leq 50
\end{array}
\][/tex]
So, the correct answer is D:
[tex]\[
\begin{array}{l}
x \geq y + 2 \\
0.50 x^2 + 1.5x + 9 y \leq 50
\end{array}
\][/tex]
