The lengths of the sides of a triangle are 10 + x 2 , 10 + x 2 and 20 - 2 x 2 . If for x = k , the area of…

The lengths of the sides of a triangle are 10+x2, 10+x2 and 20-2x2. If for x=k, the area of the triangle is maximum, then 3k2 is equal to
  1. 5
  2. 12
  3. 10
  4. 20

Solution

Area of triangle, Δ=2020-10-x220-10-x220-20+2x2

Δ=210x10-x2

For maxima, dΔdx=021010-x2-410x2=0

  x2=103

Here, d2dx2<0

Hence 3k2=10

Asked in: JEE Main 2022 (27 Jun Shift 1)

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