Chapter 6 · Application of Derivatives

Exercise 6.3 · Q6: Maximum profit when $p(x)=41-72x-18x^2$

Exercise 6.3Maximum value of a function · Section 6.4

The question

6. Find the maximum profit that a company can make, if the profit function is given by

p(x)=41-72x-18x^2.

Exam answer

$p'(x) = -72 - 36x = -36(x + 2)$ ⁠.

$p'(x) = 0$ gives $x = -2$ ⁠.

$p''(x) = -36 < 0$ for all $x$ ⁠, so $p''(-2) < 0$ ⁠: $x = -2$ is a point of maxima.

\begin{aligned} p(-2) &= 41 - 72(-2) - 18(-2)^2 \\ &= 41 + 144 - 72 = 113 \end{aligned}

The maximum profit is $113$ units.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The profit function is a quadratic with a negative $x^2$ coefficient, so its graph is a parabola opening downwards: it has exactly one highest point, where the slope $p'(x)$ is zero. Find that point, check it really is the top, and read off the profit there.

Given, to find, and the plan

Given: $p(x)=41-72x-18x^2$ ⁠. The question does not say what $x$ measures or restrict it, so, as the book does, we take $p$ as a function on all real $x$ ⁠.
To find: The maximum value of $p$ ⁠.
Plan: (1) Find the critical point from $p'(x)=0$ ⁠. (2) Confirm a local maximum with the second derivative test. (3) Prove it is the largest value of $p$ for every $x$ by completing the square. (4) Comment on the meaning of $x$ ⁠.

Basics you need

1Critical points and the second derivative test

A local maximum of a differentiable function needs $p'(c)=0$ (Theorem 2); if also $p''(c)\lt0$ ⁠, then $c$ is a point of local maxima (Theorem 4). Lesson: first and second derivative tests

2Local versus overall maximum

A local maximum is only highest nearby. To answer "the maximum profit" we must show $p(c)\ge p(x)$ for all $x$ (Definition 3; the book prints $\gt$ ⁠, read $\ge$ since $x=c$ is allowed). Lesson: maxima and minima

3Completing the square

$-18x^2-72x=-18(x^2+4x)$ $=-18\left[(x+2)^2-4\right]$ ⁠. Basics: completing the square

4Turning words into a function

Before using the answer, ask what $x$ means and which values it may take. Basics: building the function to optimise

Solution

Differentiate and find the critical point
$p'(x)=-72-36x=-36(x+2).$
$p'(x)=0$ only when $x=-2$ ⁠.

Power rule term by term: the constant $41$ gives $0$ ⁠, $-72x$ gives $-72$ ⁠, $-18x^2$ gives $-36x$ ⁠. Factoring out $-36$ makes the zero obvious. $p$ is a polynomial, differentiable everywhere, so $x=-2$ is the only critical point.

Theorem 2, Section 6.4. Lesson: tests
Second derivative test

$p''(x)=-36$ ⁠, so $p''(-2)=-36\lt0$ ⁠. By Theorem 4(i), $x=-2$ is a point of local maxima, and
$\begin{aligned} p(-2)&=41-72(-2)-18(-2)^2\\ &=41+144-72=113. \end{aligned}$
Careful with signs: $-72\times(-2)=+144$ ⁠, and $(-2)^2=4$ so $-18\times4=-72$ ⁠.

Theorem 4(i), Section 6.4.
It is the largest value for every $x$

Complete the square:
$\begin{aligned} p(x)&=41-18(x^2+4x)\\ &=41-18\left[(x+2)^2-4\right]\\ &=113-18(x+2)^2. \end{aligned}$
Since $(x+2)^2\ge0$ ⁠, $p(x)\le113$ for every real $x$ ⁠, with equality only at $x=-2$ ⁠. So the maximum value of $p$ is $113$ .

The second derivative test only promises "highest nearby". The completed square shows "highest everywhere". Check by expanding: $113-18(x^2+4x+4)$ $=113-18x^2-72x-72$ $=41-72x-18x^2$ ⁠. ✓

Definition 3, Section 6.4. Basics: completing the square
What does $x=-2$ mean?

The maximum occurs at $x=-2$ ⁠, a negative number. The question does not say what $x$ stands for, and the book's answer (113) treats $p$ as defined for every real $x$ ⁠, so that is the answer to give.

If, however, $x$ were the number of items produced, only $x\ge0$ would make sense. Then $p'(x)=-36(x+2)\lt0$ for every $x\ge0$ ⁠, so $p$ would be decreasing on $[0,\infty)$ and the largest profit would be $p(0)=41$ ⁠, at zero output.

This is the "allowed range" check that every optimisation problem needs: the critical point must lie in the range of values the variable can actually take. Here the question gives no range, so we follow the book; the note shows how the answer would change with one.

Theorem 1, Section 6.3 (negative derivative ⇒ decreasing). Basics: building the function to optimise

p(x)=41-72x-18x^2=113-18(x+2)^2

⁠: a downward parabola with its top at

(-2,113)

⁠. For

x\ge0

(right of the vertical axis) the curve only falls, starting from

p(0)=41

⁠.

Result

The maximum profit is 113 units, reached at $x=-2$ ⁠, because $p(x)=113-18(x+2)^2\le113$ for all $x$ ⁠. (If $x$ were restricted to $x\ge0$ ⁠, the largest profit would be $p(0)=41$ ⁠.)

(Textbook answer: Maximum profit = 113 unit.)

Check it with numbers

$p(-1.9)=41+136.8-64.98=112.82$ and $p(-2.1)=41+151.2-79.38=112.82$ ⁠: both just below $113$ ⁠, and equal, as the symmetry about $x=-2$ predicts ( $113-18(0.1)^2=112.82$ ⁠). ✓

$p(0)=41$ ⁠, $p(1)=41-72-18=-49$ ⁠: falling for $x\ge0$ ⁠. ✓

Common mistakes

Sign slip in $p(-2)$ ⁠. $-72(-2)=+144$ ⁠; writing $-144$ gives $-175$ ⁠, which is wrong.
Stopping at "local maximum". Show it is the largest value overall (completed square, or: a downward parabola has only one turning point).
Answering with $x=-2$ ⁠. The question asks for the maximum profit, which is $p(-2)=113$ ⁠.

What to take away

For a quadratic profit or cost function, $p'(x)=0$ gives the vertex; the sign of the $x^2$ coefficient says whether it is a maximum or a minimum.
Completing the square turns the local result into a global one in a single line.
Always check that the optimal $x$ is allowed by the meaning of the problem.

Every fact used, and where it comes from

Fact	Used in	Source
Local extremum ⇒ $p'(c)=0$ (for differentiable $p$ ⁠)	Step 1	Theorem 2, Section 6.4 (Lesson: tests)
Second derivative test	Step 2	Theorem 4, Section 6.4
Maximum value of a function (printed with $\gt$ ⁠; read $\ge$ ⁠)	Step 3	Definition 3, Section 6.4 (Lesson: maxima and minima)
Completing the square; $t^2\ge0$	Step 3	Algebra (Basics)
Negative derivative on an interval ⇒ decreasing	Step 4	Theorem 1, Section 6.3 (Lesson: increasing and decreasing)

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Exercise 6.3 · Q6: Maximum profit when p(x)=41−72x−18x2p(x)=41-72x-18x^2p(x)=41−72x−18x2

Basics you need

Solution

Differentiate and find the critical point

Second derivative test

It is the largest value for every xxx

What does x=−2x=-2x=−2 mean?

Every fact used, and where it comes from

Exercise 6.3 · Q6: Maximum profit when $p(x)=41-72x-18x^2$

It is the largest value for every $x$

What does $x=-2$ mean?