Example 37: Price $5-\frac{x}{100}$⁠, cost $\frac{x}{5}+500$⁠: sell 240 items for maximum profit

Basics you need

Solution

1. The variable and its range

   Let $x$ be the number of items sold. It is a whole number with $x \ge 0$⁠, and the price $5 - \frac{x}{100}$ can only make sense while it is not negative, that is for $x \le 500$⁠. So we work on $0 \le x \le 500$⁠, treating $x$ as a real number for the calculus and checking at the end that the answer is a whole number.

   Calculus works with a continuous variable. That is fine here as long as the best $x$ found turns out to be a whole number of items (it will).
   $5 - \frac{x}{100} \ge 0 \iff x \le 500$⁠. Basics: interval notation

2. Selling price and cost of $x$ items

   $$S(x) = \left(5 - \frac{x}{100}\right)x = 5x - \frac{x^2}{100},$$ $$C(x) = \frac{x}{5} + 500.$$
   "Price of rupees $\left(5 - \frac{x}{100}\right)$ each" is per item, so the total received for $x$ items is $x$ times it. The cost is already given for all $x$ items together.
   Total = number × price per item. Basics: building the function

3. The profit function

   $$\begin{aligned} P(x) &= S(x) - C(x) \\ &= 5x - \frac{x^2}{100} - \frac{x}{5} - 500 \\ &= \frac{24}{5}x - \frac{x^2}{100} - 500. \end{aligned}$$
   Profit is what is left after paying the cost. The two $x$-terms combine: $5x - \frac{x}{5} = \frac{25x - x}{5} = \frac{24}{5}x$⁠.
   Subtracting a bracket changes the sign of every term in it: $-\left(\frac{x}{5} + 500\right) = -\frac{x}{5} - 500$⁠.

4. Where the derivative is zero

   $$P'(x) = \frac{24}{5} - \frac{2x}{100} = \frac{24}{5} - \frac{x}{50}.$$ $$P'(x) = 0 \iff \frac{x}{50} = \frac{24}{5} \iff x = \frac{24 \times 50}{5} = 240.$$
   At a maximum inside the range the graph is flat, so $P'$ must be zero there (Theorem 2, Section 6.4). This gives the only candidate, $x = 240$⁠, which lies inside $[0, 500]$⁠.
   $\frac{d}{dx}\left(\frac{x^2}{100}\right) = \frac{2x}{100} = \frac{x}{50}$⁠; the constant $-500$ has derivative 0.

5. Second derivative test

   $$P''(x) = -\frac{1}{50}, \qquad P''(240) = -\frac{1}{50} < 0.$$

   By Theorem 4, $x = 240$ is a point of local maxima of $P$⁠.

   $P'(240) = 0$ with $P''(240) < 0$ is case (i) of the second derivative test: the profit curve is flat and bending down at $x = 240$⁠.
   Theorem 4 (Second Derivative Test), Section 6.4. Lesson: derivative tests

6. It is the largest profit over the whole range

   $P'(x) = \frac{24}{5} - \frac{x}{50}$ is positive for $x < 240$ and negative for $x > 240$⁠. So $P$ increases on $[0, 240]$ and decreases on $[240, 500]$⁠, and $P(240)$ is larger than every other value. Completing the square shows the same thing directly:

   $$\begin{aligned} P(x) &= -\frac{1}{100}\left(x^2 - 480x\right) - 500 \\ &= -\frac{(x - 240)^2}{100} + 576 - 500 \\ &= 76 - \frac{(x - 240)^2}{100}. \end{aligned}$$

   The square is never negative, so $P(x) \le 76$⁠, with equality only at $x = 240$⁠.

   The second derivative test alone guarantees only a local maximum. The question asks for the maximum profit, so we show no other number of items does better. Since $P''$ is negative everywhere, there is no other peak anywhere.
   $\frac{240^2}{100} = \frac{57600}{100} = 576$⁠. Basics: factorising and the discriminant

7. Answer the question

   $x = 240$ is a whole number, so it is an allowed number of items. The manufacturer should sell 240 items. Then each sells for $5 - \frac{240}{100} = 2.60$ rupees, he receives $240 \times 2.60 = 624$ rupees, the cost is $\frac{240}{5} + 500 = 548$ rupees, and the maximum profit is Rs 76.

   The question asks for the number of items, so that is the main answer. The profit itself (Rs 76) is a useful check: it agrees with $P(240)$ from the completed square.
   $624 - 548 = 76$⁠.

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