Chapter 7.6, Problem 38E

To determine

To find: the optimal inventory level for the model, to find the optimal profit.

Answer to Problem 38E

The optimal profit is $630000 and the optimal inventory level is 600 units of $300 profit giving model and 1200 units of $375 profit giving model.

Explanation of Solution

Given:

A manufacturer produces two models of elliptical cross-training exercise machines. The times for assembling, finishing, and packaging model $X$ are 3 hous,3 hours, and 0.8 hour, respectively. The times for model $Y$ are 4 hours, 2.5 hours, and 0.4 hour. The total times available for assembling, finishing, and packaging are 6000 hours, and 950 hours, respectively. The profit per unit are $300 for model $X$ and $375 for model $Y$ .

Calculation:

Let $x$ be model pieces produced which has the $300 and $y$ be the number of pieces produced which is having the cost of $375. So the objective function is given by:

  $p=300x+375y$

The three constraints can be converted into linear inequalities as:

  $\begin{array}{l}3x+4y\le 6000\\ 3x+2.5y\le 4200\\ 0.8x+0.4y\le 950\end{array}$

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 bounded by the constraints are shown below:

  

Now, to find the co-ordinates of the point A and B. Point A is the intersection of lines (1) and (2)

  $\begin{array}{l}3x+4y=\mathrm{6000.............}(1)\\ 3x+2.5y=\mathrm{4200.........}(2)\end{array}$

From equation (1),

  $3x=6000-4y\mathrm{............}(3)$

From equation (2),

  $3x=4200-2.5y$ …………..(4)

At the intersection point $x$ values are same. So we can equate (3) and (4) and we get , $y$ as 1200 substituting this value in equation (3), we get the value of $x$ as 600.

So the co-ordinates of the intersection point A is $\left(600,1200\right)$

Point B is the intersection of lines (5) and (6)

  $\begin{array}{l}0.8x+0.4y=\mathrm{950...........}(5)\\ 3x+2.5y=\mathrm{4200............}(6)\end{array}$

From equation (5)

  $x=1187.5-0.5y$ ………….(7)

From equation (6),

  $x=1400-\frac{2.5}{3}y$ ………………(8)

At the intersection point $x$ values are same. So we can equate (7) and (8) and we get, $y$ as 637.5. Substituting this value in equation (7), we get the value of $x$ as 868.75.

So point B is $\left(868.75,637.5\right)$ .

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

At $\left(0,0\right):p=300\left(0\right)+375\left(0\right)=0$ Minimum value of $p$

At $\left(0,1500\right):p=300\left(0\right)+375\left(1500\right)=562500$

At $\left(600,1200\right):p=300\left(600\right)+375\left(1200\right)=630000$ Maximum value of $p$

At $\left(1187.5,0\right):p=300\left(1187.5\right)+375\left(0\right)=356250$

At $\left(868.75,637.5\right):p=300\left(868.75\right)+375\left(637.5\right)=499687.5$

So the optimal profit is $630000 and the optimal inventory level is 600 units of $300 profit giving model and 1200 units of $375 profit giving model.