Math, asked by omprakashssubudhi, 8 hours ago

The cost to produce bottled spring water is given by C(x) = 16x – 63 where x is the number of thousands of bottles. The total income (revenue) from the sale of these bottles is given by the function R(x) = – x 2 + 326x – 7463.
22. Since Profit = Revenue - Cost , the profit function would be (a) – x 2 + 210x – 2400 (b) – x 2 + 210x – 7400 (c) – x 2 + 310x – 7400 (d) –x 2 – 310xproduce
23. How many bottles sold will produce the maximum profit? (a) 125 (b) 155 (c) 175 (d) 185
24. What is the maximum profit? (a) Rs 14625 (b) Rs 16625 (c) Rs 22645 (d) Rs 14685
25. What is the profit when 245 thousand bottles are sold? (a) Rs 8525 (b) Rs 9225 (c) Rs 12645 (d) Rs 10685​

Answers

Answered by juliyaton515
8

Answer:

126

Step-by-step explanation:

The cost to produce bottled spring water is given by C(x) = 16x – 63 where x is the number of thousands of bottles. The total income (revenue) from the sale of these bottles is given by the function R(x) = – x 2 + 326x – 7463.

22. Since Profit = Revenue - Cost , the profit function would be (a) – x 2 + 210x – 2400 (b) – x 2 + 210x – 7400 (c) – x 2 + 310x – 7400 (d) –x 2 – 310xproduce

23. How many bottles sold will produce the maximum profit? (a) 125 (b) 155 (c) 175 (d) 185

24. What is the maximum profit? (a) Rs 14625 (b) Rs 16625 (c) Rs 22645 (d) Rs 14685

25. What is the profit when 245 thousand bottles are sold? (a) Rs 8525 (b) Rs 9225 (c) Rs 12645 (d) Rs 10685

Answered by amitnrw
28

Given :

The cost to produce bottled spring water is given by C(x) = 16x – 63 where x is the number of thousands of bottles.

The total income (revenue) from the sale of these bottles is given by the function R(x) = – x² + 326x – 7463.

To Find :  Profit

Bottles sold will produce the maximum profit

maximum profit

profit when 245 thousand bottles are sold

Solution:

C(x) = 16x – 63

R(x) = – x² + 326x – 7463.

Profit = Revenue - Cost

p(x) = R(x) - C(x)

= – x² + 326x – 7463 -  (16x - 63)

= – x² + 310x – 7400

option  c)

Profit =  – x² + 310x – 7400

P(x) =  – x² + 310x – 7400

P'(x) = -2x + 310

P'(x) = 0  =>  -2x + 310 = 0 => x = 155

P''(x) = - 2 < 0

Hence Profit is maximum at x = 155

155   bottles sold will produce the maximum profit

option b)

maximum profit at x = 155

P(x) =  – x² + 310x – 7400

= -(155)² + 310(155) - 7400

= Rs 16625

maximum profit is Rs 16625

option b)

profit when 245 thousand bottles are sold

x = 245

profit  =  -(245)² + 310(245) - 7400

= 8525

option (a) Rs 8525

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