Chapter 7.6, Problem 42E

To determine

To find: the numbers of audits and tax returns will yield an optimal revenue and the optimal revenue.

Answer to Problem 42E

Optimal revenue is $19000 and it is obtained when there is 5 audits and 48 tax returns.

Explanation of Solution

Given:

The accounting firm in Exercise 41 lowers its charge for an audit to $1400.

Calculation:

Let $x$ be the number of audits and $y$ be the number of tax returns carried out by the firm. So the objective function is given by:

  $z=1400x+250y$

We have to find the maximum revenue.

The three constraints can be converted into linear inequalities as:

  $\begin{array}{l}60x+10y\le 780\\ 16x+4y\le 272\end{array}$

Since the number of pieces produced cannot be negative, we can get two more constraints of

  $x\ge 0$ and $y\ge 0$ .

The area determined by the constraints are shown below:

  

The point of intersection of two lines A, should be found by solving the equaton of two lines:

  $\begin{array}{l}60x+10y=\mathrm{780............}(1)\\ 16x+4y=\mathrm{272.............}(2)\end{array}$

From equation (1),

  $x=13-\frac{1}{6}y\mathrm{............}(3)$

From equation (2),

  $x=17-\frac{1}{4}y\mathrm{................}(4)$  ..........(4)

At the intersection point $x$ values are same. So we can equate (3) and (4) and we get , $y$ as 48 substituting this value in equation (4), we get the value of $x$ as 5.

So the co-ordinates of the intersection point A is $\left(5,48\right)$

At the four vertices of the region formed by the constraints the objective function has the following values:

At $\left(0,0\right):z=1400\left(0\right)+250\left(0\right)=0$

At $\left(0,68\right):z=1400\left(0\right)+250\left(68\right)=17000$

At $\left(5,48\right):z=1400\left(5\right)+250\left(48\right)=19000$ Maximum value of $z$

At $\left(13,0\right):z=1400\left(13\right)+250\left(0\right)=18200$

Optimal revenue is $19000 and it is obtained when there is 5 audits and 48 tax returns.