Question

YCorporation, a manufacturing entity, has the following profit function: \(\pi=-\$ 10,000+\$ 400 \mathrm{Q}-\$ 2 Q^{2}\) where \(\pi=\) Total profit and \(Q\) is output in units a) What will happen if output is zero? b) At what output level is profit maximized?

Short answer

a) If output is zero, the total profit is -\$10,000. b) The output level that maximizes profit is 100 units.

Step-by-step solution

1. Substitute Q = 0 in the profit function

To do this, we simply substitute Q = 0 in the profit function: \(\pi = -\$10,000 + \$400(Q) - \$2(Q^2)\)

2. Simplify the expression

When substituting Q = 0, we have: \(\pi = -\$10,000 + \$400(0) - \$2(0^2) = -\$10,000\) b) Output level that maximizes profit

3. Find the first-order derivative of the profit function

We find the derivative of the profit function with respect to Q: \(\frac{d\pi}{dQ} = 400 - 4Q\)

4. Set the first-order derivative equal to zero and solve for Q

To find the output level, set the first-order derivative equal to zero: \(400 - 4Q = 0\) Then, solve for Q: \(Q = 100\)

5. Find the second-order derivative of the profit function

We find the second-order derivative of the profit function with respect to Q: \(\frac{d^2\pi}{dQ^2} = -4\)

6. Check the sign of the second-order derivative

The second-order derivative is negative, which implies that the function is concave and has a maximum at Q = 100. Results: a) If output is zero, the total profit is -\$10,000. b) The output level that maximizes profit is 100 units.

Key concepts

First-Order Derivative

Understanding the first-order derivative is crucial when it comes to identifying profit maximization for a business. In calculus, the first-order derivative of a function represents the rate at which the function's value is changing at any given point. For a profit function, such as that of YCorporation with \(\pi=-\$ 10,000+\$ 400 \mathrm{Q}-\$ 2 Q^{2}\), finding the first-order derivative involves calculating how profit (\(\pi\)) changes with respect to changes in output (\(Q\)).

In the exercise, to find the output level that maximizes profit, we calculated the first-order derivative of the profit function: \(\frac{d\pi}{dQ} = 400 - 4Q\). To determine the profit-maximizing output, this derivative is set equal to zero. This is because the point at which the rate of change of profit (slope of \(\pi\)) is zero indicates a potential maximum or minimum. In this context, a zero derivative signals that increasing or decreasing production will not change profit, suggesting an extremum point where profit could be maximized.

Second-Order Derivative

While the first-order derivative gives us a potential maximum or minimum, the second-order derivative helps confirm the nature of this extremum. The exercise involves the second-order derivative to ascertain whether the output level found indeed maximizes profit. The second-order derivative measures the rate of change of the function's slope, essentially indicating whether the slope is getting steeper or flatter.

In the step-by-step solution, the second-order derivative calculated was \(\frac{d^2\pi}{dQ^2} = -4\), which is a constant negative value. This is a key piece of information. A negative second-order derivative implies that the slope of the profit function is decreasing as output increases, consequently confirming the presence of a concave shape. For the profit function, a concave shape at the point where the first derivative is zero means that we have found a maximum. Hence, the output level corresponding to this point is indeed where the profit is maximized.

Profit Function

The profit function is the mathematical representation of a company's profit depending on various factors, in this case, the output level. For YCorporation, the profit function is defined as \(\pi=-\$ 10,000+\$ 400 \mathrm{Q}-\$ 2 Q^{2}\). This quadratic equation implies that profit initially increases with output due to the positive \(\$400 \mathrm{Q}\) term but eventually starts decreasing as the negative \(\$2 Q^{2}\) term grows more significant. This shift is typical for profit functions since increasing production can lead to lower profits at some point due to factors like increased costs or market saturation.

The exercise asked what would happen if the output (\(Q\)) is zero. Substituting zero into the profit function gives \(\pi = -\$10,000\), indicating a loss, as fixed costs have not been covered by sales revenue, a common situation for businesses with high initial costs. Understanding the shape of a profit function is vital for businesses to decide at what point to stop increasing production to avoid decreasing profits.