Questions: The weekly revenue from the sale of x trucker hats is R(x)=-0.025 x^2+24 x dollars, and the weekly cost of producing x trucker hats is C(x)=11 x+14 dollars. Find the production level x that maximizes the weekly profit.

Solution Steps

To find the production level \( x \) that maximizes the weekly profit, we need to first define the profit function \( P(x) \) as the difference between the revenue function \( R(x) \) and the cost function \( C(x) \). Then, we find the critical points of \( P(x) \) by taking its derivative and setting it to zero. Finally, we determine which of these critical points maximizes the profit by evaluating the second derivative.

The weekly revenue from the sale of \( x \) trucker hats is given by: \[ R(x) = -0.025x^2 + 24x \]

The weekly cost of producing \( x \) trucker hats is given by: \[ C(x) = 11x + 14 \]

The profit function \( P(x) \) is the difference between the revenue and the cost: \[ P(x) = R(x) - C(x) = (-0.025x^2 + 24x) - (11x + 14) = -0.025x^2 + 13x - 14 \]

To find the production level \( x \) that maximizes the profit, we first find the critical points by taking the derivative of \( P(x) \) and setting it to zero: \[ P'(x) = \frac{d}{dx}(-0.025x^2 + 13x - 14) = -0.05x + 13 \]

Setting the first derivative to zero: \[ -0.05x + 13 = 0 \] \[ x = \frac{13}{0.05} = 260 \]

To determine whether the critical point \( x = 260 \) is a maximum, we evaluate the second derivative of \( P(x) \): \[ P''(x) = \frac{d^2}{dx^2}(-0.025x^2 + 13x - 14) = -0.05 \]

Since \( P''(x) = -0.05 \) is negative, the critical point \( x = 260 \) is a maximum.

Final Answer