Chapter 3.4, Problem 24PPS

To determine

To calculate: The number of brochures and fliers to be produced to minimize the cost.

Answer to Problem 24PPS

The number of brochures produced are 50 and fliers are 150.

Explanation of Solution

Given information:

The cost of a printing a brochure is $0.08 and that of flier cake is $0.04. A brochure requires 3 pages and a flier requires 2 pages. The manager of the company don’t want to use more than 600 pages. There is a requirement of at least 50 brochures and 150 fliers.

Calculation:

Consider the provided information that cost of a printing a brochure is $0.08 and that of flier cake is $0.04. A brochure requires 3 pages and a flier requires 2 pages. The manager of the company don’t want to use more than 600 pages. There is a requirement of at least 50 brochures and 150 fliers.

Let x denote the number of brochures and y denote the number of fliers printed.

Since, there is a requirement of at least 50 brochures and 150 fliers.

  $\begin{array}{c}x\ge 50\\ y\ge 150\end{array}$

The brochure requires 3 pages and a flier requires 2 pages. The manager of the company don’t want to use more than 600 pages.

Therefore,

  $3x+2y\le 600$.

Plot the inequalities and shade the common region $3x+2y\le 600,x\ge 50\text{ and }y\ge 150$

  

The profit function that is the cost function that is to be minimized is $f\left(x,y\right)=0.08x+0.04y$.

Substitute the vertices of the feasible region to find the point at which maximum revenue is there.

Substitute $\left(50,150\right)$ in the function $f\left(x,y\right)=0.08x+0.04y$.

  $f\left(50,150\right)=10$

Substitute $\left(50,225\right)$ in the function $f\left(x,y\right)=0.08x+0.04y$.

  $f\left(50,225\right)=13$

Substitute $\left(100,150\right)$ in the function $f\left(x,y\right)=0.08x+0.04y$.

  $f\left(100,150\right)=14$

Since, minimum cost is $10 which is when brochures produced are 50 and 150 fliers are produced.