camdengibson89
Answered

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John and his son are building a boxcar for a competition. The rules state:

1. The length of the car must be 3 inches greater than its width.
2. The width of the car must be at least 2 inches greater than the radius of the wheels.
3. They must spend no more than [tex]$50. The cost of the base is $[/tex]0.50 per square inch, and each of the 4 wheels costs $2.25 per inch of radius.

If [tex]\(x\)[/tex] represents the width of the car and [tex]\(y\)[/tex] represents the radius of the wheels, which of the following systems of inequalities can be used to determine the length and width of the car and the radii of the wheels?

A.
[tex]\[
\begin{aligned}
y & \geq z+2 \\
y & \leq \frac{N}{9} - \frac{1}{6}x - \frac{1}{9}x^2
\end{aligned}
\][/tex]

B.
[tex]\[
\begin{aligned}
y & \leq z-2 \\
y & \leq \frac{6}{9} - \frac{1}{6}z - \frac{1}{9}z^2
\end{aligned}
\][/tex]

C.
[tex]\[
\begin{aligned}
y & \geq z-2 \\
y & \leq \frac{6}{3} - \frac{1}{6}z - \frac{1}{18}z^2
\end{aligned}
\][/tex]

D.
[tex]\[
\begin{aligned}
y & \leq x-2 \\
y & \leq \frac{60}{7} - \frac{1}{6}z - \frac{1}{18}z^2
\end{aligned}
\][/tex]

Sagot :

GinnyAnswer
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]
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