Chapter 10, Problem 8P

Manufacturing Calculators A manufacturer of calculators produces two models: standard and scientific. Long-term demand for the two models mandates that the company manufacture at least 100 standard and 80 scientific calculators each day. However, because of limitations on production capacity, no more than 200 standard and 170 scientific calculators can be made daily. To satisfy a shipping contract, a total of at least 200 calculators must be shipped every day.

1. (a) If the production cost is $5 for a standard calculator and $7 for a scientific one, how many of each model should be produced daily to minimize this cost?
2. (b) If each standard calculator results in a $2 loss but each scientific one produces a $5 profit, how many of each model should be made daily to maximize profit?

To determine

To find: The number of standard and scientific calculators produced daily to minimize the cost.

Answer to Problem 8P

The company makes the 120 standard and 80 scientific calculators to minimize the production cost.

Explanation of Solution

Given:

The company manufactures at least 100 standard and 80 scientific calculators each day.

Due to limitation on production capacity, no more than 200 standard and 170 scientific calculators made daily.

The production cost on standard calculators is $\$5$ and production cost on scientific calculators is $\$7$

Calculation:

Let the number of daily production of the standard calculators is x and the number of daily production of the scientific calculators is y.

Use the given information to make the inequalities and the objective function for the feasible region.

The required information is shown in the table below.

	Standard calculators(x)	Scientific Calculators(y)
At-least production	100	80
Not more than production	200	170
Profit	$\$5$	$\$7$

The Production function is,

$P=\$5x+\$7y$,

The constraint to get the feasible region has shown below.

The production of at least 100 standard and 80 scientific calculators each day represented as,

$x\ge 100$

And,

$y\ge 80$

The production limit of the calculators is represented as,

$x\le 200$

And,

$y\le 170$

The total number of calculators must be shipped from the company at least 200.

$x+y\ge 200$

Now, take the equalities of the above inequalities,

$x=100$, $y=80$ and $x=200$

$y=170$ (1)

And,

$x+y=200$ (2)

Substitute 100 for x in the equation (1).

$\begin{array}{c}x+y=200\\ 100+y=200\\ y=200-100\\ y=100\end{array}$

Thus, the intersection point is $\left(x,y\right)=\left(100,100\right)$.

Substitute $80$ for y in the equation (5)

$\begin{array}{c}x+y=200\\ x+80=200\\ x=200-80\\ x=120\end{array}$

The intersection point is $\left(x,y\right)=\left(120,80\right)$.

Substitute 170 for y in equation (2),

$\begin{array}{c}x+y=200\\ x+170=200\\ x=200-170\\ x=30\end{array}$

Substitute $200$ for x in equation (2),

$\begin{array}{c}x+y=200\\ 200+y=200\\ y=200-200\\ y=0\end{array}$

The intersection point is $\left(x,y\right)=\left(200,0\right)$.

Now, draw the graph of the above equations,

Figure (1)

The vertices which lies in the feasible region is shown below,

$\left(100,170\right),\left(100,100\right),\left(120,80\right),\left(200,80\right),\left(200,170\right)$

Substitute the 100 for x and 170 for y in the Production function $P=\$5x+\$7y$,

$\begin{array}{c}P=5x+7y\\ =5\cdot 100+7\cdot 170\\ =500+1190\\ =1690\end{array}$

Substitute the 100 for x and 100 for y in the objective function $P=\$5x+\$7y$,

$\begin{array}{c}P=5x+7y\\ =5\cdot 100+7\cdot 100\\ =500+700\\ =1200\end{array}$

Substitute the 120 for x and 8 for y in the objective function $P=\$5x+\$7y$,

$\begin{array}{c}P=5x+7y\\ =5\cdot 120+7\cdot 80\\ =600+560\\ =1160\end{array}$

Substitute the 200 for x and 80 for y in the objective function $P=\$5x+\$7y$,

$\begin{array}{c}P=5x+7y\\ =5\cdot 200+7\cdot 80\\ =1000+560\\ =1560\end{array}$

Substitute the 200 for x and 170 for y in the objective function $P=\$5x+\$7y$,

$\begin{array}{c}P=5x+7y\\ =5\cdot 200+7\cdot 170\\ =1000+1190\\ =2190\end{array}$

So, all the satisfies these vertices are shown in the table below.

Vertices	$P=\$5x+\$7y$
$\left(100,170\right)$	$\$1690$
$\left(200,80\right)$	$\$1560$
$\left(120,80\right)$	$\$1160$ (Minimum)
$\left(200,170\right)$	$\$2190$
$\left(100,100\right)$	$\$1200$

Thus, the minimum production cost is $\$1690$ at the 120 standard calculators and 80 scientific calculators for the production function $P=\$5x+\$7y$.

To determine

To find: The number of standard and scientific calculators produced daily to maximize the profit if the standard calculators produces loss of $\$2$ and the scientific calculators produces the profit of $\$5$

Answer to Problem 8P

The company makes the 100 standard and 170 scientific calculators to maximize the profit.

Explanation of Solution

Given:

The company manufactures at least 100 standard and 80 scientific calculators each day.

Due to limitation on production capacity, no more than 200 standard and 170 scientific calculators made daily.

The each standard calculator produces a loss of $\$2$ and each scientific calculator produces a profit of $\$5$

Calculation:

Let the number of daily production of the standard calculators is x and the number of daily production of the scientific calculators is y.

Use the given information to make the inequalities and the objective function for the feasible region.

The required information is shown in the table below.

	Standard calculators(x)	Scientific Calculators(y)
At-least production	100	80
Not more than production	200	170
Profit	$-\$2$	$\$5$

The Production function is,

$M=-\$2x+\$5y$,

From the figure (1) of the part (a) keep similar constraints are shown below.

$x=100$, $y=80$ and $x=200$

$y=170$ (1)

And,

$x+y=200$ (2)

Figure (1)

The vertices which lies in the feasible region is shown below.

$\left(100,170\right),\left(100,100\right),\left(120,80\right),\left(200,80\right),\left(200,170\right)$

Substitute the 100 for x and 170 for y in the Production function $M=-\$2x+\$5y$,

$\begin{array}{c}M=-\$2x+\$5y\\ =-\$2\cdot 100+\$5\cdot 170\\ =-\$200+\$850\\ =\$650\end{array}$

Substitute the 100 for x and 100 for y in the objective function $M=-\$2x+\$5y$,

$\begin{array}{c}M=-\$2x+\$5y\\ =-\$2\cdot 100+\$5\cdot 100\\ =-\$200+\$500\\ =\$300\end{array}$

Substitute the 120 for x and 8 for y in the objective function $M=-\$2x+\$5y$,

$\begin{array}{c}M=-\$2x+\$5y\\ =-\$2\cdot 120+\$5\cdot 80\\ =-\$240+\$400\\ =\$160\end{array}$

Substitute the 200 for x and 80 for y in the objective function $M=-\$2x+\$5y$,

$\begin{array}{c}M=-\$2x+\$5y\\ =-\$2\cdot 200+\$5\cdot 80\\ =-\$400+\$400\\ =0\end{array}$

Substitute the 200 for x and 170 for y in the objective function $M=-\$2x+\$5y$,

$\begin{array}{c}M=-\$2x+\$5y\\ =-\$2\cdot 200+\$5\cdot 170\\ =-\$400+\$850\\ =\$450\end{array}$

So, all the satisfies these vertices are shown in the table below.

Vertices	$M=-\$2x+\$5y$
$\left(100,170\right)$	$\$650$( Maximize)
$\left(200,80\right)$	$\$0$
$\left(120,80\right)$	$\$160$
$\left(200,170\right)$	$\$450$
$\left(100,100\right)$	$\$300$

Thus, the maximum production cost is $\$650$ at the 100 standard calculators and 170 scientific calculators for the production function $M=-\$2x+\$5y$.