Question

a lunch stand makes a $.75 profit on each chef's salad and $1.20 profit on each caesar salad. On a typical weekday, it sells between 40 and 60 chefs salads and between 35 and 50 caesar salads. the total number sold has never exceed 100 salads. how many of each type should be prepared in order to maximize profit?

Answer 1

Answer:

50 Chef's salads and 50 Caesar salads should be prepared in order to maximize profit.

Step-by-step explanation:

Suppose, the number of Chef's salad is $x$ and the number of Caesar salad is $y$

On a typical weekday, it sells between 40 and 60 Chefs salads and between 35 and 50 Caesar salads.

So, the two constraints are:  $40\leq x\leq 60$ and  $35\leq y\leq 50$

The total number sold has never exceed 100 salads. So, another constraint will be:   $x+y\leq 100$

According to the graph of the constraints, the vertices of the common shaded region are:  $(40,35), (60,35), (60,40), (50,50)$ and $(40,50)$   (Refer to the attached image for the graph)

The lunch stand makes a $.75 profit on each Chef's salad and $1.20 profit on each Caesar salad. So, the profit function will be:  $P=0.75x+1.20y$

For  (40, 35) ,   $P=0.75(40)+1.20(35)=72$

For  (60, 35) ,   $P=0.75(60)+1.20(35)=87$

For  (60, 40) ,   $P=0.75(60)+1.20(40)=93$

For  (50, 50) ,   $P=0.75(50)+1.20(50)=97.5$ (Maximum)

For  (40, 50) ,   $P=0.75(40)+1.20(50)=90$

Profit will be maximum when $x=50$ and $y=50$

Thus, 50 Chef's salads and 50 Caesar salads should be prepared in order to maximize profit.