8. Consider the profit functionπ = −5Q2 + 200Q + 10where Q is the quantity of goods sold.(a) Compute dπdQ and d2πdQ2 .(b) Use part (a) to find the maximum profit, and write for which value of Qthe maximum is attaine
Question
- Consider the profit functionπ = −5Q2 + 200Q + 10where Q is the quantity of goods sold.(a) Compute dπdQ and d2πdQ2 .(b) Use part (a) to find the maximum profit, and write for which value of Qthe maximum is attaine
Solution
(a) To find the first derivative of the profit function π with respect to Q (dπ/dQ), we apply the power rule of differentiation, which states that the derivative of x^n, where n is any real number, is n*x^(n-1).
The derivative of -5Q^2 is -10Q (using the power rule), the derivative of 200Q is 200 (since the derivative of any constant times x is just the constant), and the derivative of 10 is 0 (since the derivative of a constant is 0).
So, dπ/dQ = -10Q + 200.
To find the second derivative of the profit function π with respect to Q (d²π/dQ²), we again apply the power rule to our first derivative.
The derivative of -10Q is -10 and the derivative of 200 is 0.
So, d²π/dQ² = -10.
(b) To find the maximum profit, we set the first derivative equal to zero and solve for Q.
-10Q + 200 = 0 -10Q = -200 Q = 20
The second derivative test tells us that if the second derivative at a point is negative, then the function has a local maximum at that point. Since d²π/dQ² = -10, which is negative, we know that the profit function has a local maximum at Q = 20.
Therefore, the maximum profit is attained when Q = 20. To find the maximum profit, we substitute Q = 20 into the profit function:
π = -5(20)^2 + 200*20 + 10 π = -2000 + 4000 + 10 π = 2010
So, the maximum profit is 2010.
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