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Soloha48 [4]
3 years ago
7

Kelly runs a restaurant that sells two kinds of desserts. Kelly knows the restaurant must make at least 25 dozen and at most 45

dozen White Chocolate Blizzards. The restaurant must also make at least 2 dozen and at most 15 dozen Mint Breezes. Each dozen of White Chocolate Blizzards takes 5 ounces, of flour, while each dozen of Mint Breezes requires 9 ounces of flour. The restaurant only has 315 ounces of flour available. If a dozen White Chocolate Blizzards generate $2.92 in income, and a dozen Mint Breezes generate $2.11, how many dozens of the desserts should Kelly have the restaurant make to maximize income? Express your answer as an ordered pair in the form (c,m) where c represents dozens of White Chocolate Blizzards and m represents dozens of Mint Breezes.

Mathematics
1 answer:
MissTica3 years ago
8 0

Answer:

  (c, m) = (45, 10)

Step-by-step explanation:

A dozen White Chocolate Blizzards generate more income and take less flour than a dozen Mint Breezes, so production of those should clearly be maximized. Making 45 dozen Blizzards does not use all the flour, so the remaining flour can be used to make Breezes.

Maximum Blizzards that can be made: 45 dz. Flour used: 45×5 oz = 225 oz.

The remaining flour is ...

  315 oz -225 oz = 90 oz

This is enough for (90 oz)/(9 oz/dz) = 10 dozen Mint Breezes. This is in the required range of 2 to 15 dozen.

Kelly should make 45 dozen White Chocolate Blizzards and 10 dozen Mint Breezes: (c, m) = (45, 10).

__

In the attached graph, we have reversed the applicable inequalities so the feasible region shows up white, instead of shaded with 5 different colors. The objective function is the green line, shown at the point that maximizes income. (c, m) ⇔ (x, y)

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Number of Women, n(W)=3

Total Sample, n(S)=24+3=27

Since you cannot appoint the same person twice, the probabilities are <u>without replacement.</u>

(a)Probability that both appointees are men.

P(MM)=\dfrac{24}{27}X \dfrac{23}{26}=\dfrac{552}{702}\\=\dfrac{92}{117}

(b)Probability that one man and one woman are appointed.

To find the probability that one man and one woman are appointed, this could happen in two ways.

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P(One man and one woman are appointed)=P(MW)+P(WM)

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(c)Probability that at least one woman is appointed.

The probability that at least one woman is appointed can occur in three ways.

  • A man is appointed first and a woman is appointed next.
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P(at least one woman is appointed)=P(MW)+P(WM)+P(WW)

P(WW)=\dfrac{3}{27}X \dfrac{2}{26}=\dfrac{6}{702}

In Part B, P(MW)+P(WM)=\frac{8}{39}

Therefore:

P(MW)+P(WM)+P(WW)=\dfrac{8}{39}+\dfrac{6}{702}\\$P(at least one woman is appointed)=\dfrac{25}{117}

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