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Consider the following constrained maximization problem: \\[ \begin{array}{ll} \text { maximize } & y=x_{1}+5 \ln x_{2} \\ \text { subject to } & k-x_{1}-x_{2}=0 \end{array} \\] where \(k\) is a constant that can be assigned any specific value. a show that if \(k=10,\) this problem can be solved as one involving only equality constraints. b. Show that solving this problem for \(k=4\) requires that \\[ x_{1}=-1 \\] c. If the \(x\) 's in this problem must be non-negative, what is the optimal solution when \(k=4 ?\) (This problem may be solved cither intuitively or using the methods outlined in the chapter.) d. What is the solution for this problem when \(k=20 ?\) What do you conclude by comparing this solution with the solution for part (a)?

Short Answer

Expert verified
In summary, we solved the constrained maximization problem involving the variables \(x_1\) and \(x_2\) and a constraint relating them with a given constant \(k\). We rewrote the problem as a single-variable problem, allowing us to analyze how different values of \(k\) affect the constraint and the objective function. We also considered non-negativity requirements for the variables and explored the impacts of these constraints on possible solutions for different values of \(k\). For \(k = 10\), the non-negativity constraint does not impose any issues on the problem. However, for \(k = 4\), we must ensure that both \(x_1\) and \(x_2\) are non-negative to find the optimal solution. When solving the problem for \(k = 20\), we found that the range of possible values for \(x_2\) expands, affecting the location of the maximum point in the problem.

Step by step solution

01

Rewrite the constrained problem as a single-variable problem

First, let's examine the constraint equation: \\[ k - x_1 - x_2 = 0 \\] We can rewrite the equation to make either \(x_1\) or \(x_2\) the subject of the equation. Let's make \(x_1\) the subject: \\[ x_1 = k - x_2 \\] Now, we can replace \(x_1\) in the objective function with the expression we found: \\[ y = (k - x_2) + 5 \ln x_2 \\]
02

Analyze the constraint with different values of \(k\)

We can investigate how the constraint affects the problem by looking at the values of \(k\) provided in the exercise and plugging them into the constraint equation: a) For \(k = 10\), \\[ x_1 = 10 - x_2 \\] b) For \(k = 4\), \\[ x_1 = 4 - x_2 \\]
03

Solve the single-variable problem for different values of \(k\)

We will now solve the single-variable problem using the objective function for each value of \(k\): a) For \(k = 10\), the objective function becomes: \\[ y = (10 - x_2) + 5 \ln x_2 \\] b) For \(k = 4\), the objective function becomes: \\[ y = (4 - x_2) + 5 \ln x_2 \\]
04

Find possible solutions for different values of \(k\)

Now, we will analyze the objective function for the given values of \(k\) and verify if non-negativity is a requirement. Additionally, we will investigate the impact of this constraint on the solution when it is necessary. a) For \(k = 10\), we see that \(x_1 = 10 - x_2\). Since \(x_1\) can only be non-negative (as per the exercise's instructions), we can deduce that \(x_2 \leq 10\). As a result, the non-negativity constraint is not an issue in this case. b) For \(k = 4\), we see that \(x_1 = 4 - x_2\). Since \(x_1\) must be non-negative, we deduce that \(x_2 \leq 4\). We can find a point in the objective function where \(x_1\) is negative (for example, when \(x_2 > 4\)), leading to the result: \\[ x_1 = -1 \\] c) If the \(x\)s in this problem must be non-negative, and \(k = 4\), we can find the optimal solution by looking at the constraint and the objective function. When \(k = 4\), the constraint becomes \(x_1 = 4 - x_2\). We can plug this into our objective function, which then becomes: \\[ y = (4 - x_2) + 5 \ln x_2 \\] To solve this problem, we must ensure that \(x_1\) and \(x_2\) are non-negative, meaning that \(0 \leq x_1 \leq 4\) and \(0 \leq x_2 \leq 4\). Detecting the maximum point in this domain will provide the optimal solution. d) When \(k = 20\), the constraint is now \(x_1 = 20 - x_2\). Since \(x_1\) must be non-negative, we deduce that \(x_2 \leq 20\). Comparing this solution with the solution for part (a) reveals that as the value of \(k\) increases, the range of possible \(x_2\) values expands as well, thereby affecting the location of the maximum point in the problem.

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Most popular questions from this chapter

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