Chapter 6 · Application of Derivatives

Exercise 6.3 · Q14: $x+y=60$ ⁠, largest $xy^3$ ⁠: $x=15$ ⁠, $y=45$

Exercise 6.3Optimisation · Section 6.4

The question

Find two positive numbers $x$ and $y$ such that $x + y = 60$ and $xy^3$ is maximum.

Exam answer

From $x + y = 60$ ⁠, $y = 60 - x$ ⁠, where $0 < x < 60$ ⁠. Let

f(x) = xy^3 = x(60 - x)^3

By the product rule,

\begin{aligned} f'(x) &= (60 - x)^3 - 3x(60 - x)^2 \\ &= (60 - x)^2(60 - 4x) \\ &= 4(60 - x)^2(15 - x) \end{aligned}

For $0 < x < 60$ ⁠, $f'(x) = 0$ gives $x = 15$ ⁠.

In $(0, 15)$ ⁠, $f'(x) > 0$ ⁠. In $(15, 60)$ ⁠, $f'(x) < 0$ ⁠. So $f$ is greatest at $x = 15$ ⁠.

$x = 15$ and $y = 45$ ⁠.

The answer key lists 45, 15 without saying which is $x$ ⁠; the maximum needs $x = 15$ ⁠, $y = 45$ ⁠.

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

The one idea this problem is built on

Use $x + y = 60$ to write $xy^3$ as a function of $x$ alone, on the range $0 < x < 60$ that keeps both numbers positive. Factorising its derivative shows a single sign change, from $+$ to $-$ ⁠, so the critical point gives the largest value on the whole range. The power 3 on $y$ means $y$ deserves the bigger share: the answer is not "equal halves".

Given, to find, and the plan

Given: $x > 0$ ⁠, $y > 0$ and $x + y = 60$ ⁠.
To find: The $x$ and $y$ for which $xy^3$ is largest.
Plan: Put $y = 60 - x$ ⁠; find the range of $x$ ⁠; let $f(x) = x(60 - x)^3$ ⁠; differentiate with the product and chain rules and factorise; locate the critical point in the range; test it, and use the sign of $f'$ to show it is the largest value of all.

Basics you need

1Building the function and its range

Replace $y$ using the condition, and write down exactly which values of $x$ are allowed. Basics: building the function to optimise

2Product rule and chain rule

$(uv)' = u'v + uv'$ ⁠; $\frac{d}{dx}(60 - x)^3$ $= 3(60 - x)^2 \cdot (-1)$ (outer $t^3$ ⁠, inner $60 - x$ ⁠). Basics: the chain rule

3Sign of a product

A square is positive (when nonzero), so the sign of $(60 - x)^2(15 - x)$ is the sign of $15 - x$ ⁠. Basics: sign of a product

4Local versus largest

Increasing up to $c$ and decreasing after $c$ on the allowed range means $f(c)$ is the largest value there (Theorem 1, Section 6.3). Lesson: optimisation problems

Solution

One variable and its allowed range

From $x + y = 60$ ⁠, $y = 60 - x$ ⁠. Both numbers must be positive:
$x > 0 \quad\text{and}\quad 60 - x > 0,$
so the allowed range is
$0 < x < 60.$
Every allowed pair $(x, y)$ corresponds to exactly one $x$ in $(0, 60)$ ⁠, and every $x$ in $(0, 60)$ gives an allowed pair. So maximising over this open interval is the same as maximising over all allowed pairs. The ends 0 and 60 are excluded because "positive" means strictly greater than 0.

Interval notation: $(0, 60)$ means $0 < x < 60$ ⁠. Basics: interval notation
The function to maximise
$f(x) = x\,y^3 = x(60 - x)^3,\qquad 0 < x < 60.$
Substituting $y = 60 - x$ turns a two-variable expression into a one-variable function that we can differentiate.

Keep it factored: the derivative will factorise neatly. Basics: building the function
Differentiate and factorise

Product rule with $u = x$ ⁠, $v = (60 - x)^3$ ⁠; chain rule for $v' = 3(60 - x)^2 \cdot (-1)$ ⁠:
$\begin{aligned} f'(x) &= (60 - x)^3 - 3x(60 - x)^2 \\ &= (60 - x)^2\,[(60 - x) - 3x] \\ &= (60 - x)^2\,(60 - 4x) \\ &= 4(60 - x)^2\,(15 - x). \end{aligned}$
Both terms contain $(60 - x)^2$ ⁠, so take it out as a common factor; what is left, $60 - 4x$ ⁠, equals $4(15 - x)$ ⁠. The factored form shows at once where $f'$ is zero and what sign it has.

Chain rule: outer function $t^3$ (derivative $3t^2$ ⁠), inner function $t = 60 - x$ (derivative $-1$ ⁠). Basics: the chain rule
Critical point in the range
$f'(x) = 0 \iff x = 60 \ \text{ or } \ x = 15.$
$x = 60$ is not in the allowed range (it would make $y = 0$ ⁠), so the only critical point is
$x = 15.$
$f$ is a polynomial, differentiable everywhere, so critical points are only where $f' = 0$ ⁠. The algebra also produces $x = 60$ ⁠, but the range rules it out.

Critical point: Note after Theorem 2, Section 6.4. Lesson: maxima and minima
Sign of $f'$ ⁠: the largest value on the whole range

For $0 < x < 60$ ⁠, $(60 - x)^2 > 0$ ⁠. So $f'(x) = 4(60 - x)^2(15 - x)$ has the sign of $15 - x$ ⁠:
- for $0 < x < 15$ ⁠: $f'(x) > 0$ ⁠, so $f$ is increasing on $(0, 15]$ ⁠;
- for $15 < x < 60$ ⁠: $f'(x) < 0$ ⁠, so $f$ is decreasing on $[15, 60)$ ⁠.
Hence $f(x) < f(15)$ for every other $x$ in $(0, 60)$ ⁠. (The second derivative agrees: $f''(x) = -12(60 - x)(30 - x)$ ⁠, so $f''(15) = -12 \times 45 \times 15$ $= -8100 < 0$ ⁠, a local maximum by Theorem 4.)

$f(x) = x(60 - x)^3$ on $0 < x < 60$ (vertical scale in millions). One peak, at $x = 15$ ⁠; the open dots at the ends are excluded. The rose dot shows the swapped choice $x = 45$ ⁠, which is far worse.

The first derivative changes from $+$ to $-$ only once in the whole range, so $f$ climbs all the way to $x = 15$ and descends all the way after it. That proves the global maximum, which the second derivative test alone would not.

Theorem 1, Section 6.3 (and the Remark after it, for intervals whose end points are excluded). Lesson: increasing and decreasing
The numbers, and the largest value

$x = 15$ and $y = 60 - 15 = 45$ ⁠. The largest value is
$\begin{aligned} f(15) &= 15 \times 45^3 \\ &= 15 \times 91125 = 1366875. \end{aligned}$
Answer in terms of the named numbers: here the order matters because $x$ and $y$ play different roles in $xy^3$ ⁠.

$45^2 = 2025$ and $2025 \times 45 = 91125$ ⁠. Basics: building the function

Result

$x = 15$ and $y = 45$ ; then $xy^3 = 1366875$ ⁠, the largest possible value. (Textbook answer: 45, 15. The key lists the same two numbers, 45 and 15, without saying which one is $x$ ⁠. Read in the order $x = 45$ ⁠, $y = 15$ they would make $xy^3$ only $45 \times 15^3 = 151875$ ⁠, so the key's pair agrees with this page only as $y = 45$ ⁠, $x = 15$ ⁠.)

Check it with numbers

$f(14) = 14 \times 46^3 = 1362704$ and $f(16) = 16 \times 44^3 = 1362944$ ⁠, both a little below $f(15) = 1366875$ ⁠. Working with $y$ instead gives the same answer: $g(y) = (60 - y)y^3$ has $g'(y) = 4y^2(45 - y)$ ⁠, zero at $y = 45$ ⁠. A computer search over $x = 0.001, 0.002, \dots, 59.999$ finds the largest value at $x = 15$ ⁠.

Common mistakes

Splitting 60 equally. Equal halves maximise $xy$ ⁠, not $xy^3$ ⁠; the cube rewards a larger $y$ ⁠.
Swapping $x$ and $y$ ⁠. $x = 45$ ⁠, $y = 15$ gives only 151875. The key's "45, 15" names the two numbers without labelling them; the larger one must be $y$ ⁠.
Losing the minus sign in the chain rule. $\frac{d}{dx}(60 - x)^3 = -3(60 - x)^2$ ⁠.
Keeping $x = 60$ ⁠. It makes $y = 0$ ⁠, which is not positive.

What to take away

For $x + y = s$ with $x, y > 0$ (and positive powers $m, n$ ⁠), the product $x^m y^n$ is largest when $x : y = m : n$ ⁠; here $1 : 3$ ⁠, so $x = 15$ ⁠, $y = 45$ ⁠.
Factor out the common power in the derivative; then the sign of $f'$ is easy to read on the whole range.
The same pattern with $x^2y^5$ is Question 15; the plain product is Question 13.

Every fact used, and where it comes from

Fact	Used in	Source
Writing the quantity in one variable with its range	Steps 1, 2	Section 6.4 examples (Lesson: optimisation problems)
Product rule; chain rule	Step 3	Chapter 5 (Basics: the chain rule)
Critical point	Step 4	Note after Theorem 2, Section 6.4 (Lesson: maxima and minima)
Sign of $f'$ decides increasing or decreasing	Step 5	Theorem 1 and the Remark after it, Section 6.3 (Lesson: increasing and decreasing)
Second derivative test	Step 5	Theorem 4, Section 6.4 (Lesson: derivative tests)
Answer 45, 15	Result	Answers to Exercise 6.3, Question 14

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Exercise 6.3 · Q14: x+y=60x+y=60x+y=60⁠, largest xy3xy^3xy3⁠: x=15x=15x=15⁠, y=45y=45y=45

Basics you need

Solution

One variable and its allowed range

The function to maximise

Differentiate and factorise

Critical point in the range

Sign of f′f'f′⁠: the largest value on the whole range

The numbers, and the largest value

Every fact used, and where it comes from

Exercise 6.3 · Q14: $x+y=60$ ⁠, largest $xy^3$ ⁠: $x=15$ ⁠, $y=45$

Sign of $f'$ ⁠: the largest value on the whole range