The weekly revenue for a product is given by ​R(x)equals288 x minus 0.054 x squared​, and the weekly cost is ​C(x)equals10 comma 000 plus 144 x minus 0.108 x squared plus 0.00003 x cubed​, where x is the number of units produced and sold.

Question

​(a) How many units will give the maximum​ profit?
​(b) What is the maximum possible​ profit?

Answer 1

To find the maximum profit, we first need to determine the profit function \(P(x)\), which is defined as the difference between the revenue function \(R(x)\) and the cost function \(C(x)\).

Given:

• Revenue: \[ R(x) = 288x - 0.054x^2 \]

• Cost: \[ C(x) = 10,000 + 144x - 0.108x^2 + 0.00003x^3 \]

Profit Function:

The profit function \(P(x)\) is given by: \[ P(x) = R(x) - C(x) \] Substituting in the expressions for \(R(x)\) and \(C(x)\): \[ P(x) = (288x - 0.054x^2) - (10,000 + 144x - 0.108x^2 + 0.00003x^3) \]

Now, let's simplify this: \[ P(x) = 288x - 0.054x^2 - 10,000 - 144x + 0.108x^2 - 0.00003x^3 \] Combining like terms: \[ P(x) = (288 - 144)x + (-0.054 + 0.108)x^2 - 0.00003x^3 - 10,000 \] \[ P(x) = 144x + 0.054x^2 - 0.00003x^3 - 10,000 \]

(a) Finding the Maximum Profit:

To find the maximum profit, we need to find the critical points by taking the derivative of \(P(x)\) and setting it equal to zero. \[ P'(x) = 144 + 0.108x - 0.00009x^2 \] Setting \(P'(x) = 0\): \[ 144 + 0.108x - 0.00009x^2 = 0 \] Rearranging gives: \[ 0.00009x^2 - 0.108x - 144 = 0 \]

Next, we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 0.00009\), \(b = -0.108\), and \(c = -144\).

Calculating the discriminant: \[ b^2 - 4ac = (-0.108)^2 - 4(0.00009)(-144) \] \[ = 0.011664 + 0.05184 = 0.063504 \]

Now we can find \(x\): \[ x = \frac{-(-0.108) \pm \sqrt{0.063504}}{2(0.00009)} \] Calculating \( \sqrt{0.063504} \): \[ \sqrt{0.063504} \approx 0.252 \] \[ x = \frac{0.108 \pm 0.252}{0.00018} \] Calculating both potential values: \[ x_1 = \frac{0.360}{0.00018} \approx 2000 \] \[ x_2 = \frac{-0.144}{0.00018} \text{ (not relevant since } x \text{ cannot be negative)} \]

Thus, the number of units that maximizes profit is: \[ \boxed{2000} \]

(b) Calculating the Maximum Profit:

Now substituting \(x = 2000\) into the profit function \(P(x)\): \[ P(2000) = 144(2000) + 0.054(2000^2) - 0.00003(2000^3) - 10,000 \] Calculating each term:

• \(144(2000) = 288000\)
• \(0.054(2000^2) = 0.054 \times 4000000 = 216000\)
• \(0.00003(2000^3) = 0.00003 \times 8000000000 = 240000\)

Now substitute these values back into the profit formula: \[ P(2000) = 288000 + 216000 - 240000 - 10,000 \] \[ = 288000 + 216000 - 240000 - 10,000 = 254000 \]

Thus, the maximum possible profit is: \[ \boxed{254000} \]