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ryzh [129]
3 years ago
10

A manufacturer produces two types of bottled coffee drinks: cappuccinos and cafés au lait. Each bottle of cappuccino requires 6

ounces of coffee and 2 ounces of milk and earns a profit of $0.40. Each bottle of café au lait requires 4 ounces of coffee and 4 ounces of milk and earns a profit of $0.50. The manufacturer has 720 ounces of coffee and 400 ounces of milk available for production each day. To meet demand, the manufacturer must produce at least 80 coffee drinks each day. Let x = the number of cappuccino bottles and y = the number of café au lait bottles.

Mathematics
2 answers:
arlik [135]3 years ago
8 0
<span>x + y => 80
</span>6x + 4y <= 720
<span>2x + 4y <= 400</span>
gladu [14]3 years ago
8 0

Solution:

It is given that ,  x = the number of cappuccino bottles and y = the number of café au lait bottles.

Let p ounces of coffee is mixed with q ounces of milk to make a bottle of cappuccino.

1.  x = 6 p + 2 q

2.  y = 4 p + 4 q

3.  x + y ≥ 80

Amount of coffee possessed by manufacturer= 720 ounces

Amount of milk possessed by manufacturer= 400 ounces

Equation (1) gives

→6 p = 720

→→p=120 ounces(Number of bottles of coffee only)

→→→2 q = 400

→ q = 200 ounces(Number of bottles of milk only)

Maximum number of cappuccino bottles = 120

Equation (2) gives

→→4 p = 720

→ p = 180 ounces((Number of bottles of coffee only)

→→→ 4 q = 400

→→q = 100 ounces(Number of bottles of milk only)

Maximum number of Cafe au lait bottles = 100

So, x = The Maximum number of cappuccino bottles = 120

y=The Maximum number of café au lait bottles.=100  

Also the manufacturer can try different combinations of Cappuccino and cafe au lait bottles of coffee to manufacture more than 80 bottles in a day. for example, (56,24), (34,46).....

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