Chapter 6 · Application of Derivatives

Exercise 6.3 · Q15: $x+y=35$ ⁠, largest $x^2y^5$ ⁠: $x=10$ ⁠, $y=25$

Exercise 6.3Optimisation · Section 6.4

The question

Find two positive numbers $x$ and $y$ such that their sum is 35 and the product $x^2y^5$ is a maximum.

Exam answer

From $x + y = 35$ ⁠, $y = 35 - x$ ⁠, where $0 < x < 35$ ⁠. Let

f(x) = x^2y^5 = x^2(35 - x)^5

By the product rule,

\begin{aligned} f'(x) &= 2x(35 - x)^5 \\ &\quad - 5x^2(35 - x)^4 \\ &= x(35 - x)^4(70 - 7x) \\ &= 7x(35 - x)^4(10 - x) \end{aligned}

For $0 < x < 35$ ⁠, $f'(x) = 0$ gives $x = 10$ ⁠.

In $(0, 10)$ ⁠, $f'(x) > 0$ ⁠. In $(10, 35)$ ⁠, $f'(x) < 0$ ⁠. So $f$ is greatest at $x = 10$ ⁠.

$x = 10$ and $y = 25$ ⁠.

The answer key lists 25, 10 without saying which is $x$ ⁠; the maximum needs $x = 10$ ⁠, $y = 25$ ⁠.

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

The one idea this problem is built on

Use the sum to write the product in terms of $x$ alone on the range $0 < x < 35$ ⁠. The derivative factorises into pieces whose signs are obvious on that range; it changes sign only once, from $+$ to $-$ ⁠, so the single critical point gives the largest product of all.

Given, to find, and the plan

Given: $x > 0$ ⁠, $y > 0$ and $x + y = 35$ ⁠.
To find: The $x$ and $y$ for which $x^2y^5$ is largest.
Plan: Put $y = 35 - x$ ⁠, find the range of $x$ ⁠, let $f(x) = x^2(35 - x)^5$ ⁠; differentiate (product and chain rules), take out the common factors, find the critical point inside the range, and read the sign of $f'$ on the whole range.

Basics you need

1Building the function and its range

Replace $y$ using the condition and state the allowed values of $x$ ⁠. Basics: building the function to optimise

2Product rule and chain rule

$(uv)' = u'v + uv'$ ⁠; $\frac{d}{dx}(35 - x)^5$ $= 5(35 - x)^4 \cdot (-1)$ ⁠. Basics: the chain rule

3Sign of a product

On $0 < x < 35$ ⁠: $x > 0$ and $(35 - x)^4 > 0$ ⁠, so only the remaining factor decides the sign. Basics: sign of a product

4Local versus largest

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

Solution

One variable and its allowed range

From $x + y = 35$ ⁠, $y = 35 - x$ ⁠. Both numbers positive means $x > 0$ and $35 - x > 0$ ⁠:
$0 < x < 35.$
Each allowed pair matches exactly one $x$ in $(0, 35)$ and vice versa, so maximising over this interval covers every allowed pair. The ends are excluded because the numbers must be strictly positive.

$35 - x > 0 \iff x < 35$ ⁠. Basics: interval notation
The function to maximise
$f(x) = x^2y^5 = x^2(35 - x)^5,\qquad 0 < x < 35.$
After substituting $y = 35 - x$ the product depends on $x$ only.

Keep the factored form; the derivative will share these factors. Basics: building the function
Differentiate and factorise

Product rule with $u = x^2$ ⁠, $v = (35 - x)^5$ ⁠; chain rule gives $v' = -5(35 - x)^4$ ⁠:
$\begin{aligned} f'(x) &= 2x(35 - x)^5 - 5x^2(35 - x)^4 \\ &= x(35 - x)^4\,[2(35 - x) - 5x] \\ &= x(35 - x)^4\,(70 - 7x) \\ &= 7x(35 - x)^4\,(10 - x). \end{aligned}$
Both terms contain $x$ and $(35 - x)^4$ ⁠; taking them out leaves $2(35 - x) - 5x$ $= 70 - 7x = 7(10 - x)$ ⁠.

Chain rule: outer $t^5$ (derivative $5t^4$ ⁠), inner $t = 35 - x$ (derivative $-1$ ⁠). Basics: the chain rule
Critical point in the range
$f'(x) = 0 \iff x = 0,\ x = 35 \ \text{ or } \ x = 10.$
Only $x = 10$ lies in $0 < x < 35$ ⁠.

$x = 0$ and $x = 35$ would make one number zero, which is not positive. $f$ is a polynomial, so there are no points where it fails to be differentiable.

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

For $0 < x < 35$ ⁠, $7x > 0$ and $(35 - x)^4 > 0$ ⁠, so $f'(x)$ has the sign of $10 - x$ ⁠:
- for $0 < x < 10$ ⁠: $f'(x) > 0$ ⁠, so $f$ is increasing on $(0, 10]$ ⁠;
- for $10 < x < 35$ ⁠: $f'(x) < 0$ ⁠, so $f$ is decreasing on $[10, 35)$ ⁠.
So $f(x) < f(10)$ for every other allowed $x$ ⁠: by the first derivative test (Theorem 3) $x = 10$ is a point of local maxima, and the sign pattern makes it the largest value on the whole range.

$f(x) = x^2(35 - x)^5$ on $0 < x < 35$ ⁠, vertical scale in units of $10^8$ ⁠. The single peak is at $x = 10$ ⁠. The rose dot is the swapped choice $x = 25$ ⁠, which gives a much smaller product.

Here the first derivative test is the natural choice: the factored $f'$ shows the sign change directly, while $f''$ would be a long computation. The same sign information also gives the global conclusion.

First Derivative Test, Theorem 3, Section 6.4; Theorem 1, Section 6.3. Lesson: derivative tests
The numbers, and the largest value

$x = 10$ and $y = 35 - 10 = 25$ ⁠. The largest value is
$\begin{aligned} f(10) &= 10^2 \times 25^5 \\ &= 100 \times 9765625 \\ &= 976562500. \end{aligned}$
The question names $x$ (squared) and $y$ (to the fifth), so say which number is which.

$25^5 = 25^2 \times 25^2 \times 25$ $= 625 \times 625 \times 25 = 9765625$ ⁠. Basics: building the function

Result

$x = 10$ and $y = 25$ ; then $x^2y^5 = 976562500$ ⁠, the largest possible value. (Textbook answer: 25, 10. The key lists the same two numbers, 25 and 10, without saying which one is $x$ ⁠. Read in the order $x = 25$ ⁠, $y = 10$ they would make $x^2y^5$ only $625 \times 10^5 = 62500000$ ⁠, so the key's pair agrees with this page only as $y = 25$ ⁠, $x = 10$ ⁠.)

Check it with numbers

$f(9) = 81 \times 26^5 = 962391456$ and $f(11) = 121 \times 24^5 = 963477504$ ⁠, both below $f(10) = 976562500$ ⁠. The ratio rule (see "What to take away") predicts $x : y = 2 : 5$ ⁠, that is $x = \frac{2}{7} \times 35 = 10$ ⁠. A computer search over $x = 0.001, 0.002, \dots, 34.999$ finds the largest value at $x = 10$ ⁠.

Common mistakes

Expanding $(35 - x)^5$ ⁠. It works but is long and error-prone; take out common factors instead.
Dropping the $-1$ from the chain rule, which gives $2(35 - x) + 5x$ and no critical point in the range.
Swapping the numbers. $x = 25$ ⁠, $y = 10$ gives 62500000, far less. The key's "25, 10" names the two numbers without labelling them; the larger one must be $y$ ⁠.
Keeping $x = 0$ or $x = 35$ ⁠. They make $f' = 0$ but are not allowed (the numbers must be positive), and there $f = 0$ ⁠, the opposite of a maximum.

What to take away

For $x + y = s$ with $x, y > 0$ ⁠, $x^m y^n$ is largest when $x : y = m : n$ ⁠: here $2 : 5$ ⁠, so $x = 10$ ⁠, $y = 25$ ⁠.
Factor the derivative completely; on the allowed range most factors have a fixed sign, and one factor decides everything.
Compare Question 14 ( $xy^3$ ⁠, sum 60).

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)
First derivative test	Step 5	Theorem 3, Section 6.4 (Lesson: derivative tests)
Sign of $f'$ decides increasing or decreasing	Step 5	Theorem 1 and the Remark after it, Section 6.3 (Lesson: increasing and decreasing)
Answer 25, 10	Result	Answers to Exercise 6.3, Question 15

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Exercise 6.3 · Q15: x+y=35x+y=35x+y=35⁠, largest x2y5x^2y^5x2y5⁠: x=10x=10x=10⁠, y=25y=25y=25

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 · Q15: $x+y=35$ ⁠, largest $x^2y^5$ ⁠: $x=10$ ⁠, $y=25$

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