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kirill115 [55]
4 years ago
9

2. A lunch stand makes s.75 profit on each chef's salad and $1.20 profit on each Caesar salad. On a typical weekday, it sells be

tween 40 and 60 chef's salads and between 35 and 50 Caesar salads. The total number sold has never exceeded 100 salads. How many of each type should be prepared in order to maximize profit?
Could you make a graph as the answer?

Mathematics
1 answer:
kow [346]4 years ago
8 0

Let's let

<u>x = the number of chef salads, x>=0</u>

<u>y = the number of Caesar salads, y>=0</u>

The constrains are:

40 <= x <= 60

35 <= y <= 50

x + y <= 100

The objective function here is F(x, y) = 0.75x + 1.20y

The corner points are (40, 35),  (60, 35), (60, 40), (50, 50) and (40, 50).

F (40, 35) = 0.75*40 + 1.20*35 = $72

F (60, 35) = 0.70*60 + 1.20*35 = $84

F (60, 40) = 0.75*60 + 1.20*40 = $93

F (50, 50) = 0.75*50 + 1.20*50 = $97.50

F (40, 50) = 0.75*40 + 1.20*50 = $90

Thus, we conclude to maximize the profit 50 Chef and 50 Caesar salads should be prepared.

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44 cents

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a unit price is how much something costs per one

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4/9=.4444444....

this to the nearest hundredth is .44 or 44 cents

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we are have to know for which value or input units are these functions at maximum which translates to for how many units is the revenue maximum and for how many same units is our cost minimum.

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18. The perimeter is simply √3 + √3 + √3 + √3 or 4√3cm, since the perimeter is just all sides added together. You could add the decimal numbers together using a calculator, which I'm not sure if you're supposed to do in your class.

The area is just width times length, so √3 • √3 = 3cm².

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The area would be √5 • (9 - √5), which leaves you with (9√5 - 5)ft².

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