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3.5 · Q2

Q.The sum of two positive numbers is 16. Find the numbers, if the product of the squares is to be maximum.

Puducherry CbseNCERTSubjective· 3mImportance★★★★★
60% · 52/87 Questions
✓ Free question

Write one number as xx and the other as 16−x16-x; maximising the product of their squares P=x2(16−x)2P=x^2(16-x)^2 gives x=8x=8, so both numbers equal 88.

Maximise P(x)P(x) by solving P′(x)=0P'(x)=0 and confirming P′′(x)<0P''(x)<0 (or a sign change of P′P') at that point.

  1. Let the two positive numbers be xx and 16−x16-x (their sum is 1616).
  2. Product of their squares: P=x2(16−x)2P=x^2(16-x)^2, with 0<x<160<x<16.
  3. Differentiate (product rule): P′=2x(16−x)2+x2⋅2(16−x)(−1)=2x(16−x)[(16−x)−x]P'=2x(16-x)^2+x^2\cdot 2(16-x)(-1)=2x(16-x)\big[(16-x)-x\big].
  4. Simplify: P′=2x(16−x)(16−2x)P'=2x(16-x)(16-2x).
  5. Set P′=0⇒x=0, x=16,P'=0\Rightarrow x=0,\ x=16, or x=8x=8. The endpoints x=0,16x=0,16 give P=0P=0 (minimum), so the maximum is at x=8x=8.
  6. At x=8x=8: the other number is 16−8=816-8=8, and P=82⋅82=64⋅64=4096P=8^2\cdot 8^2=64\cdot 64=4096 (maximum).
✓Final answer

The numbers are 8\mathbf{8} and 8\mathbf{8}, giving the maximum product of squares =4096=4096.

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