The revenue function for a product is r =600q 0.5q2 and the cost funct

1.	The revenue function for a product is r =600q 0.5q2 and the cost function is c=1500 +140q-4q2+5q3. Determine the profit function and the value of q for which profit is maximum
Answer» <P>Answer: p = - 5q³ + 3.5q² + 460q - 1500 q = 5.79 Explanation: The revenue function for a PRODUCT is r =600q - 0.5q2 and the cost function is c=1500 +140q-4q2+5q3. Determine the profit function and the value of q for which profit is MAXIMUM revenue r =600q - 0.5q² Cost c = 1500 + 140q - 4q² + 5q³ Profit = Revenue - cost => p = r - c => p = (600q - 0.5q²) - (1500 + 140q - 4q² + 5q³) => p = - 5q³ + 3.5q² + 460q - 1500 dp/dq = -15q² + 7.5q + 460 equating dp/dq = 0 => -15q² + 7.5q + 460 = 0 dividing by -5 => 3q² - 1.5q - 92 =0 => q = (1.5 ± √(1.5² - 4(3)(-92)) )/6 => q = (1.5 ± 33.26)/6 => q = 5.79 , - 5.29 dp/dq = -15q² + 7.5q + 460 d²p/dq² = -30q + 7.5 Putting q = 5.79 d²p/dq² < 0 => q = 5.79 will give maximum profit

The revenue function for a product is r =600q 0.5q2 and the cost function is c=1500 +140q-4q2+5q3. Determine the profit function and the value of q for which profit is maximum

Answer»

<P>Answer:

p = - 5q³ + 3.5q² + 460q - 1500

q = 5.79

Explanation:

The revenue function for a PRODUCT is r =600q - 0.5q2 and the cost function is c=1500 +140q-4q2+5q3. Determine the profit function and the value of q for which profit is MAXIMUM

revenue r =600q - 0.5q²

Cost c = 1500 + 140q - 4q² + 5q³

Profit = Revenue - cost

=> p = r - c

=> p = (600q - 0.5q²) - (1500 + 140q - 4q² + 5q³)

=> p = - 5q³ + 3.5q² + 460q - 1500

dp/dq = -15q² + 7.5q + 460

equating dp/dq = 0

=> -15q² + 7.5q + 460 = 0

dividing by -5

=> 3q² - 1.5q - 92 =0

=> q = (1.5 ± √(1.5² - 4(3)(-92)) )/6

=> q = (1.5 ± 33.26)/6

=> q = 5.79 , - 5.29

dp/dq = -15q² + 7.5q + 460

d²p/dq² = -30q + 7.5

Putting q = 5.79 d²p/dq² < 0

=> q = 5.79 will give maximum profit

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