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2025 PEA Q18

Find non-negative x,yx,y with x+y=16x + y = 16 minimising x3+y3x^3 + y^3.

Reveal answer and solution

Answer

A

Solution

  1. 1

    Substitute y=16−xy = 16 - x:

  2. 2
    g(x)=x3+(16−x)3,x∈[0,16]. g(x) = x^3 + (16-x)^3, \quad x \in [0,16].
  3. 3
    g′(x)=3x2−3(16−x)2=3(x−(16−x))(x+(16−x))=3(2x−16)(16). g'(x) = 3 x^2 - 3(16-x)^2 = 3(x - (16-x))(x + (16-x)) = 3(2x-16)(16).
  4. 4

    g′(x)=0  ⟺  x=8g'(x) = 0 \iff x = 8. Second derivative g′′(x)=6x+6(16−x)=96>0g''(x) = 6x + 6(16-x) = 96 > 0, so x=8x=8 is a minimum.

Answer structure / marking notes

Although x3+y3x^3+y^3 is unbounded above on the line x+y=16x+y=16, on [0,16][0,16] the maxima are at the endpoints; the minimum is at the symmetric point.

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Content note

Imported from public/resources/isi/msqe/solutions/pea/2025/ISI_MSQE_PEA_2025_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.