Question

A firm uses two inputs, K and L to manufacture final product. The price per unit of these inputs are sh. 20 and sh. 4 respectively. What combination of inputs should a firm use to maximize output given that the budget is fixed at sh. 390?

Answer 1

ANSWER on Question #78911 – Math – Linear Algebra

QUESTION

A firm uses two inputs, $K$ and $L$ to manufacture final product. The price per unit of these inputs are sh. 20 and sh. 4 respectively. What combination of inputs should a firm use to maximize output given that the budget is fixed at sh. 390?

SOLUTION

Let the manufacture of $K$ produce $x$ units of the finished product, and the manufacture of $L - y$ units.

Then, the total number of products produced is

$$\text{Units}(x, y) = x + y$$

According to the problem: the manufacture $K$ pays sh. 20 for each unit and the manufacture $L - sh. 4$.

Then, the total amount of money spent is

$$\text{Cost}(x, y) = 20x + 4y$$

By the condition of the task, the amount of money spent is fixed and equal

$$\text{Cost}(x, y) = 390 \rightarrow \boxed{20x + 4y = 390}$$

Now we can write the given problem in symbolic form:

$$\left\{ \begin{array}{l} \max \bigl(\text{Cost}(x + y)\bigr) = \max(x + y) \\ \quad 20x + 4y = 390 \\ \quad \quad x \geq 0 \\ \quad \quad y \geq 0 \\ \quad \quad x, y \in \mathbb{N} \end{array} \right.$$

Then,

$$20x + 4y = 390 \rightarrow 4y = 390 - 20x \mid \div (4) \rightarrow y = \frac{390}{4} - \frac{20x}{4} \rightarrow \boxed{y = \frac{195}{2} - 5x}$$ $$\max(x, y) = x + y = \frac{195}{2} - 5x + x = \frac{195}{2} - 4x \rightarrow \boxed{\max(x, y) = \frac{195}{2} - 4x}$$

As we can see,

$$\max (x, y) = \frac {195}{2} - 4x - \text{decreasing linear function } \forall x \geq 0$$

Then,

$$\max (x, y) = \frac {195}{2} - 4x \rightarrow \frac {195}{2} - \text{for } x = 0.$$ $$\left\{\begin{array}{l} 20x + 4y = 390 \\ x = 0 \end{array}\right. \rightarrow 4y = 390 \rightarrow y = \frac {195}{2} = 97.5$$

But, $y$ is the number of products manufactured, it can only be an integer, so $y = 97$.

Conclusion,

$$\max \bigl (Cost(x, y) \bigr) = \max (x + y) = 97$$

ANSWER:

$$\max_{\left\{\begin{array}{l} 20x + 4y = 390 \\ x, y \geq 0 \end{array}\right.}(x + y) = 97$$

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