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Chapter 6 · Application of Derivatives

Exercise 6.3 · Q16: Sum 16, least sum of cubes: 88 and 88

Exercise 6.3Optimisation · Section 6.4
The question

Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Exam answer

Let the numbers be xx and 16−x16 - x⁠, where 0<x<160 < x < 16⁠. The sum of their cubes is

S(x)=x3+(16−x)3S(x) = x^3 + (16 - x)^3 S′(x)=3x2−3(16−x)2=3x2−3(256−32x+x2)=96x−768=96(x−8)\begin{aligned} S'(x) &= 3x^2 - 3(16 - x)^2 \\ &= 3x^2 - 3(256 - 32x + x^2) \\ &= 96x - 768 = 96(x - 8) \end{aligned}

S′(x)=0S'(x) = 0 gives x=8x = 8⁠.

S′′(x)=96>0S''(x) = 96 > 0 for all xx⁠, so S′′(8)>0S''(8) > 0⁠: x=8x = 8 is a point of minima.

The numbers are 88 and 88⁠.

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

The one idea this problem is built on

Write the sum of cubes as a function of one number, xx⁠, on the range 0<x<160 < x < 16⁠. The cubes x3x^3 cancel when expanded, leaving a quadratic in xx that opens upward, so there is exactly one lowest point. Calculus finds it, and the sign of the derivative proves it is the least value on the whole range.

Given, to find, and the plan
Given
Two positive numbers with sum 16.
To find
The numbers for which the sum of their cubes is least.
Plan
Numbers xx and 16−x16 - x with 0<x<160 < x < 16⁠; S(x)=x3+(16−x)3S(x) = x^3 + (16 - x)^3⁠; solve S′(x)=0S'(x) = 0⁠; second derivative test; sign of S′S' on the whole range for the global minimum.

Basics you need

1Building the function and its range

Use the sum to write the second number as 16−x16 - x⁠, and keep both numbers positive. Basics: building the function to optimise

2Difference of two squares

a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b)⁠; it turns x2−(16−x)2x^2 - (16 - x)^2 into a simple product. Basics: factorising

3Second derivative test (Theorem 4, Section 6.4)

If f′(c)=0f'(c) = 0 and f′′(c)>0f''(c) > 0⁠, then cc is a point of local minima. Lesson: derivative tests

4Local versus least

Decreasing up to cc and increasing after cc means f(c)f(c) is the least value on the range (Theorem 1, Section 6.3). Lesson: optimisation problems

Solution

  1. One variable and its allowed range

    Let one number be xx⁠; the other is 16−x16 - x⁠. Both are positive, so x>0x > 0 and 16−x>016 - x > 0⁠:

    0<x<16.0 < x < 16.
    Every allowed pair corresponds to exactly one xx in (0,16)(0, 16)⁠. The ends are excluded because "positive" means strictly greater than 0.
    16−x>0  ⟺  x<1616 - x > 0 \iff x < 16⁠. Basics: interval notation

  2. The sum of cubes

    S(x)=x3+(16−x)3,0<x<16.S(x) = x^3 + (16 - x)^3,\qquad 0 < x < 16.
    "The sum of whose cubes": cube each number and add.
    Model first, then differentiate. Basics: building the function

  3. Differentiate and factorise

    Chain rule: ddx(16−x)3\frac{d}{dx}(16 - x)^3 =3(16−x)2⋅(−1)= 3(16 - x)^2 \cdot (-1)⁠. So

    S′(x)=3x2−3(16−x)2=3 [x−(16−x)] [x+(16−x)]=3 (2x−16)(16)=96 (x−8).\begin{aligned} S'(x) &= 3x^2 - 3(16 - x)^2 \\ &= 3\,[x - (16 - x)]\,[x + (16 - x)] \\ &= 3\,(2x - 16)(16) \\ &= 96\,(x - 8). \end{aligned}
    x2−(16−x)2x^2 - (16 - x)^2 is a difference of two squares, a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b) with a=xa = x⁠, b=16−xb = 16 - x⁠. Then a+b=16a + b = 16 and a−b=2x−16=2(x−8)a - b = 2x - 16 = 2(x - 8)⁠, so 3×2×16=963 \times 2 \times 16 = 96⁠.
    Chain rule: outer t3t^3⁠, inner t=16−xt = 16 - x with derivative −1-1⁠. Basics: the chain rule

  4. Critical point and second derivative test

    S′(x)=0  ⟺  x=8,S′′(x)=96.S'(x) = 0 \iff x = 8,\qquad S''(x) = 96.

    x=8x = 8 lies in (0,16)(0, 16)⁠, and S′′(8)=96>0S''(8) = 96 > 0⁠, so by the second derivative test (Theorem 4) x=8x = 8 is a point of local minima of SS⁠.

    S′(x)=96x−768S'(x) = 96x - 768⁠, so its derivative is the constant 96. Positive second derivative means the graph bends upward: a valley.
    Theorem 4(ii), Section 6.4. Lesson: derivative tests

  5. It is the least value on the whole range

    S′(x)=96(x−8)S'(x) = 96(x - 8) is negative for 0<x<80 < x < 8 and positive for 8<x<168 < x < 16⁠. So SS is decreasing on (0,8](0, 8] and increasing on [8,16)[8, 16)⁠, and S(x)>S(8)S(x) > S(8) for every other allowed xx⁠.

    Expanding confirms it without calculus. Since (16−x)3(16 - x)^3 =4096−768x+48x2−x3= 4096 - 768x + 48x^2 - x^3⁠,

    S(x)=48x2−768x+4096=48(x−8)2+1024 ≥ 1024.\begin{aligned} S(x) &= 48x^2 - 768x + 4096 \\ &= 48(x - 8)^2 + 1024 \ \ge\ 1024. \end{aligned}
    4 8 12 16 1000 2000 3000 4000 x S(x) least: S(8) = 1024 → 4096, never reached S′ < 0 S′ > 0
    S(x)=x3+(16−x)3S(x) = x^3 + (16 - x)^3 on 0<x<160 < x < 16⁠: a U-shaped curve, symmetric about x=8x = 8⁠, with its lowest point S(8)=1024S(8) = 1024⁠. The open end dots are excluded, so values near 4096 are approached but never reached.
    The second derivative test is a local statement. The sign of S′S' (falling all the way to 8, rising all the way after) is what makes 1024 the least value on the entire range.
    48(x−8)2=48x2−768x+307248(x - 8)^2 = 48x^2 - 768x + 3072⁠, and 3072+1024=40963072 + 1024 = 4096⁠. Basics: sign of a product

  6. Answer the question

    x=8x = 8⁠, and the other number is 16−8=816 - 8 = 8⁠. The least sum of cubes is 83+83=512+512=10248^3 + 8^3 = 512 + 512 = 1024⁠.

    The question asks for the numbers; the least value is given as a check.
    83=5128^3 = 512⁠. Basics: building the function
Result

The numbers are 8 and 8; the least possible sum of their cubes is 1024. (Textbook answer: 8, 8.)

Check it with numbers

73+93=343+729=10727^3 + 9^3 = 343 + 729 = 1072⁠, 43+123=64+1728=17924^3 + 12^3 = 64 + 1728 = 1792 and 13+153=1+3375=33761^3 + 15^3 = 1 + 3375 = 3376⁠: all above 1024, and each exceeds 1024 by 48(x−8)248(x - 8)^2 (for x=7x = 7⁠: 48×1=4848 \times 1 = 48⁠). A computer search over x=0.001,0.002,…,15.999x = 0.001, 0.002, \dots, 15.999 finds the least value 1024 at x=8x = 8⁠.

Common mistakes
  1. Writing −3(16−x)2-3(16 - x)^2 as +3(16−x)2+3(16 - x)^2⁠. The inner function 16−x16 - x has derivative −1-1⁠.
  2. Expanding (16−x)3(16 - x)^3 wrongly. (a−b)3=a3−3a2b+3ab2−b3(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3⁠: here 4096−768x+48x2−x34096 - 768x + 48x^2 - x^3⁠.
  3. Looking for a maximum as well. On 0<x<160 < x < 16 there is none: SS approaches 4096 near the ends but never reaches it, because 0 is not a positive number.
What to take away
  1. For a fixed sum of two positive numbers, the sum of their cubes (like the sum of their squares) is least when the numbers are equal.
  2. A difference of squares in the derivative often factorises instantly.
  3. The book's version with squares is Example 22 (sum 15, least sum of squares).

Every fact used, and where it comes from

FactUsed inSource
Writing the quantity in one variable with its rangeSteps 1, 2Section 6.4, Example 22 (Lesson: optimisation problems)
Chain rule; difference of two squaresStep 3Chapter 5; algebra (Basics: factorising)
Second derivative testStep 4Theorem 4, Section 6.4 (Lesson: derivative tests)
Sign of f′f' decides increasing or decreasingStep 5Theorem 1 and the Remark after it, Section 6.3 (Lesson: increasing and decreasing)
Cube of a differenceStep 5Algebra
Answer 8, 8ResultAnswers to Exercise 6.3, Question 16
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