Q. 29 - Precalculus Enhanced with Graphing Utilities

Question

Product Design An entrepreneur is having a design group produce at least six samples of a new kind of fastener that he wants to market. It costs \($9.00$ to produce each metal fastener and \)$4.00$ to produce each plastic fastener. He wants to have at least two of each version of the fastener and needs to have all the samples $24$ hours (hr) from now. It takes $4$ hr to produce each metal sample and $2$ hr to produce each plastic sample. To minimize the cost of the samples, how many of each kind should the entrepreneur order? What will be the cost of the samples? 

Step-by-Step Solution

Answer

He should produce $2$ metal fasteners and $4$ plastic fasteners and it would cost $$34$.

Step 1. Given Information

The cost to produce each metal fastener and each plastic fastener is $$9.00$ and $$4.00$ respectively.

He wants to have at least two of each version of the fastener. 

It takes $4$ hr to produce each metal sample and $2$ hr to produce each plastic sample. 

Step 2. Form the inequalities

Let he produce $x$ number of metal fasteners and $y$ plastic fasteners.

He wants to have at least two of each version of the fastener and at least a total of six. So

$$\begin{gathered} x\ge 2 \\ y\ge 2 \\ x+y\ge 6 \end{gathered}$$

Time to produce $x$ metal fasteners and $y$ plastic fasteners is given as $4x+2y$ hours and it should be less than $24$.

$4x+2y\le 24$

Step 3. Form the expression that needs to be minimized

The cost of producing one metal fastener is $$9.00$ and one plastic fastener is $$4.00$. So the cost of producing $x$ metal fasteners and $y$ plastic fasteners is given by the expression $9x+4y$.

And the cost needs to be minimized, so $C=9x+4y$ needs to be minimized.

Step 4. Graph the inequalities

The inequalities are graphed and the feasible region has been shaded as 

Step 5. Find the cost at each corner point

The feasible region is marked by four boundary points and they are $\left(4,2\right),\left(5,2\right),\left(2,8\right),\left(2,4\right)$.

Now the cost at each point is found as

$\left(x,y\right)$	$C=9x+4y$
$\left(4,2\right)$	$9·4+4·2=44$
$\left(5,2\right)$	$9·5+4·2=53$
$\left(2,8\right)$	$9·2+4·8=50$
$\left(2,4\right)$	$9·2+4·4=34$

It can be seen that the minimum cost is $$34$ and it happens at $\left(2,4\right)$ that is when $2$ metal fasteners and $4$ plastic fasteners are produced.