Chapter 6.4, Problem 68E

(a)

To determine

To Find: How many of each model should be assembled each day to provide maximum profit.

Answer to Problem 68E

  $3$ of model $28$ and $10$ of model $74$ .

Explanation of Solution

Given: Profit on model $28$ is $\$100$ and on model $74$ is $\$60$ .

Assuming that $x$ represent the model $28$ cards and $y$ represents the model $74$ cards.

Then

Using the given informtaion.

The set of equations becomes

  $\begin{array}{l}30x+15y\le 240\\ 20x+30y\le 360\\ 12x+10y\le 480\\ x\ge 0\\ y\ge 0\end{array}$

The graph of the equations is shown below.

  

Now

  $\begin{array}{l}P(x,y)=100x+60y\\ P(0,0)=100(0)+60(0)=0\\ P(0,12)=100(0)+60(12)=720\\ P(3,10)=100(3)+60(10)=900\\ P(0,8)=100(0)+60(8)=480\end{array}$

From here it is seen that

  $3$ of model $28$ and $10$ of model $74$ .

(b)

To determine

To Find: The maximum daily profit.

Answer to Problem 68E

  $\$900$

Explanation of Solution

Given: Profit on model $28$ is $\$100$ and on model $74$ is $\$60$ .

  $\begin{array}{l}P(x,y)=100x+60y\\ P(0,0)=100(0)+60(0)=0\\ P(0,12)=100(0)+60(12)=720\\ P(3,10)=100(3)+60(10)=900\\ P(0,8)=100(0)+60(8)=480\end{array}$

The maximum daily profit is $\$900$ .