Certainly! Let's solve this problem step-by-step, given the constraints for the variables [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
1. Identify the constraints:
- [tex]\(0 \leq x \leq 60\)[/tex]
- [tex]\(4x + 5y \leq 400\)[/tex]
- [tex]\(y \geq 0\)[/tex]
2. Interpret the constraints:
- The first constraint [tex]\(0 \leq x \leq 60\)[/tex] means that [tex]\(x\)[/tex] must be between 0 and 60, inclusive.
- The second constraint [tex]\(4x + 5y \leq 400\)[/tex] is a linear inequality that combines [tex]\(x\)[/tex] and [tex]\(y\)[/tex].
- The third constraint [tex]\(y \geq 0\)[/tex] indicates that [tex]\(y\)[/tex] must be non-negative.
3. Find the boundary values for [tex]\(y\)[/tex] given the constraints:
- Let's look at the second equation [tex]\(4x + 5y \leq 400\)[/tex]. We need to solve this inequality for [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex].
4. Examine [tex]\(y\)[/tex] at the minimum value of [tex]\(x\)[/tex]:
- When [tex]\(x = 0\)[/tex]:
[tex]\[
4(0) + 5y \leq 400 \implies 5y \leq 400 \implies y \leq \frac{400}{5} \implies y \leq 80
\][/tex]
So, the maximum value of [tex]\(y\)[/tex] when [tex]\(x\)[/tex] is 0 is 80.
5. Examine [tex]\(y\)[/tex] at the maximum value of [tex]\(x\)[/tex]:
- When [tex]\(x = 60\)[/tex]:
[tex]\[
4(60) + 5y \leq 400 \implies 240 + 5y \leq 400 \implies 5y \leq 160 \implies y \leq \frac{160}{5} \implies y \leq 32
\][/tex]
So, the maximum value of [tex]\(y\)[/tex] when [tex]\(x\)[/tex] is 60 is 32.
6. Summarize the results:
- When [tex]\(x = 0\)[/tex], [tex]\(y\)[/tex] can be as large as 80.
- When [tex]\(x = 60\)[/tex], [tex]\(y\)[/tex] can be as large as 32.
So, the boundary values for [tex]\(y\)[/tex] are 80 when [tex]\(x\)[/tex] is 0, and 32 when [tex]\(x\)[/tex] is 60. This gives us the final values [tex]\(y = 80\)[/tex] and [tex]\(y = 32\)[/tex] for the respective [tex]\(x\)[/tex] values.
