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andrew-mc [135]
3 years ago
14

An animal food company must produce 200kg. Of a

Mathematics
1 answer:
Sati [7]3 years ago
3 0

Answer:

Minimum cost = 1200$

x_1 = 80~kg\\x_2 = 120~kg      

Step-by-step explanation:

We are given the following information in the question:

An animal food company must produce 200 kg

x_1 + x_2 = 200\\x_1 \geq 0\\x_2 \geq 0

No more than 80 kg Of first ingredient can be used and at

least 60 kg Of second ingredient must be used.

x_1 \leq 80\\x_2 \geq 60

Cost of ingredient x_1 = Rs 3 per kg

Cost of ingredient x_2 = Rs 8 per kg

Total cost = 3x_1 + 8x_2

We have to minimize this cost.

Then, we can write the following inequalities:

6x + 8y \geq 208\\x \geq 8\\y \geq 8\\

The corner points as evaluated from graph are: (0,200) and (80,120).

C(0,200) = 1600$

C(80,120) = 1200$

Hence, by corner point theorem, the minimum cost would be 1200$ when 80 kg of first ingredient is used and 120 kg of second ingredient.

The attached image shows the graph.

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Second question: 4080

Step-by-step explanation:

Here is the Sum formula:

S_{n}=\frac{n}{2}(a_{1}+a_{n})

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You can find the last term (a_{n}) by substituting k for 20 into the formula.

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S_{20}= 1270

Answer to the first question: 1270

Next question:

Even though the given formula uses n as the variable, this problem works the same way as the previous one.

Substitute n in the formula for k, which is 5 to find the first term: 14(5)+29=99

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S_{20}= \frac{20}{2} (99+309)

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S_{20}= 10(408)

S_{20}=4080

Answer to the second question: 4080

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3 years ago
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