For the following function, one zero is given. Find all the others
1 answer:
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Answer:
<em>50 Chef's salads and 50 Caesar salads should be prepared in order to maximize profit.</em>
Step-by-step explanation:
Suppose, the number of Chef's salad is and the number of Caesar salad is
On a typical weekday, it sells between 40 and 60 Chefs salads and between 35 and 50 Caesar salads.
So, the two constraints are: and
The total number sold has never exceed 100 salads. So, another constraint will be:
According to the graph of the constraints, the vertices of the common shaded region are: and
<em>(Refer to the attached image for the graph)</em>
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:
For (40, 35) ,
For (60, 35) ,
For (60, 40) ,
For (50, 50) , <u><em>(Maximum)</em></u>
For (40, 50) ,
Profit will be maximum when and
Thus, 50 Chef's salads and 50 Caesar salads should be prepared in order to maximize profit.
Answer:
Harry buys 6 apples and 5 oranges at the store.
Step-by-step explanation:
Given:
Cost of each apples = $0.8
Cost of each oranges = $0.95
Total money spend =$9.55
Number of fruits bought= 11
We need to find the number of apples bought and number of oranges were bought.
Solution:
Let the number of apples bought be 'a'.
Let the number of oranges were bought be 'o'.
Number of fruits bought= 11
So we can say that;
Number of fruits bought is equal to sum number of apples bought and number of oranges were bought.
framing in equation form we get;
⇒ equation 1
Now we can say that;
Total money spend is equal to number of apples bought multiplied by Cost of each apples plus number of oranges were bought multiplied by Cost of each oranges.
framing in equation form we get;
⇒ equation 2
Now Multiplying equation 1 by 0.8 we get;
⇒ equation 3
Subtracting equation 3 from equation 2 we get;
Dividing both side by 0.15 we get;
Substituting the value of o in equation 1 we get;
Hence Harry buys 6 apples and 5 oranges at the store.