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NARA [144]
3 years ago
5

© MP.1 Make Sense and Persevere

Mathematics
1 answer:
Andreas93 [3]3 years ago
4 0

Answer:

September = 2697

Step-by-step explanation:

Given

June = 1,898

July = 2,643

August = 2,287

Expected = 9,525

Required

Number of cans in September

First, we need to determine the total collected can from July to August.

Total = 1898 + 2643 + 2287

Total = 6828

September is then calculated as follows:

September = 9525 - 6828

September = 2697

Hence:

2697 cans are still needed

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Since profit can't be negative, the production level that'll maximize profit is approximately equal to 220.

<h3>How to find the production level that'll maximize profit?</h3>

The cost function, C(x) is given by 12000 + 400x − 2.6x² + 0.004x³ while the demand function, P(x) is given by 1600 − 8x.

Next, we would differentiate the cost function, C(x) to derive the marginal cost:

C(x) = 12000 + 400x − 2.6x² + 0.004x³

C'(x) = 400 − 5.2x + 0.012x².

Also, revenue, R(x) = x × P(x)

Revenue, R(x) = x(1600 − 8x)

Revenue, R(x) = 1600x − 8x²

Next, we would differentiate the revenue function to derive the marginal revenue:

R'(x) = 1600 - 8x

At maximum profit, the marginal revenue is equal to the marginal cost:

1600 - 8x = 400 − 5.2x + 0.012x

1600 - 8x - 400 + 5.2x - 0.012x² = 0

1200 - 2.8x - 0.012x² = 0

0.012x² + 2.8x - 1200 = 0

Solving by using the quadratic equation, we have:

x = 220.40 or x = -453.73.

Since profit can't be negative, the production level that'll maximize profit is approximately equal to 220.

Read more on maximized profit here: brainly.com/question/13800671

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