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CAT QuantmediumMaxima and Minima

Maxima and Minima: Finding the Best Value

Updated 22 October 2026

Five questions on the greatest and least values of an expression.

Question 1 · Multiple choice (+3 / −1)

If x and y are positive real numbers with x + y = 12, what is the greatest possible value of x²y?

Show the solution

Answer: B. 256

  1. Split x into two equal halves so that the three parts have a fixed sum: x/2 + x/2 + y = 12.
  2. By AM ≥ GM, (x/2 × x/2 × y) ≤ (12 ÷ 3)³ = 64, so x²y/4 ≤ 64 and x²y ≤ 256.
  3. Equality needs x/2 = y, i.e. x = 8 and y = 4: 64 × 4 = 256.
  4. Using x = y = 6 gives 216, a common slip: the parts must be equal, not x and y.
Question 2 · Type in the answer (+3 / 0)

What is the least value of |x − 1| + |x − 4| + |x − 9| for real x? Enter the number.

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Answer: 8

  1. |x − a| is the distance from x to a, so the expression is the total distance from x to 1, 4 and 9.
  2. Total distance to a set of points is least at their median, here x = 4.
  3. At x = 4: 3 + 0 + 5 = 8.
  4. Moving right of 4 adds distance to two points and saves it on one; moving left does the same. So 8 is the minimum.
Question 3 · Type in the answer (+3 / 0)

What is the least value of 4ˣ + 4¹⁻ˣ for real x? Enter the number.

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Answer: 4

  1. Both terms are positive, and their product is 4ˣ × 4¹⁻ˣ = 4, which is fixed.
  2. By AM ≥ GM, 4ˣ + 4¹⁻ˣ ≥ 2√4 = 4.
  3. Equality when 4ˣ = 4¹⁻ˣ, i.e. x = 1 − x, so x = ½: 2 + 2 = 4.
Question 4 · Multiple choice (+3 / −1)

A farmer has 120 m of fencing to enclose a rectangular plot along a straight river. No fence is needed along the river. What is the largest area that can be enclosed?

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Answer: B. 1,800 m²

  1. Let each side perpendicular to the river be x m. The side parallel to the river is then 120 − 2x m.
  2. Area A = x(120 − 2x) = 120x − 2x².
  3. A quadratic with a negative x² term peaks at its vertex: x = 120 ÷ (2 × 2) = 30.
  4. A = 30 × 60 = 1,800 m².
  5. A square (40 × 40 = 1,600 m²) is best only when all four sides are fenced; 3,600 m² comes from fencing a 60 × 60 square with no river.
Question 5 · Multiple choice (+3 / −1)

If x and y are positive real numbers with xy = 36, what is the least value of x + 4y?

Show the solution

Answer: B. 24

  1. The product x × 4y = 4 × 36 = 144 is fixed.
  2. By AM ≥ GM, x + 4y ≥ 2√144 = 24.
  3. Equality when x = 4y: with xy = 36 this gives 4y² = 36, so y = 3 and x = 12.
  4. Check: 12 + 12 = 24. Taking x = y = 6 gives 30, and x = 1, y = 36 gives far more.

Full worked solution

  1. AM ≥ GM: for positive numbers with a fixed sum, the product is greatest when they are equal; with a fixed product, the sum is least when they are equal.
  2. Split terms so the parts you make equal have a fixed sum or product, e.g. write x²y as (x/2)(x/2)y.
  3. A quadratic ax² + bx + c has its turning point at x = −b/2a: a maximum when a < 0, a minimum when a > 0.
  4. A sum of distances |x − a| + |x − b| + … is least at the median of a, b, …