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:
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:
Option 1 -
5
Option 2 -
8
Option 3 -
10
Option 4 -
12
-
1 Answer
-
Correct Option - 3
Detailed Solution:CD = √ (10+x²)² – (10–x²)² = 2√10|x|
Area
= 1/2 × CD × AB = 1/2 × 2√10|x| (20–2x²)
=> 10 – x² = 2x
3x² = 10
x = k
3k² = 10
Similar Questions for you
option (C) is incorrect, there will be minima.

absolute minimum
absolute maximum = 3
If f(x) has maximum value at x = 1 then
……..(i)
……..(ii)
From (i) and (ii) we get
OP2 = x2 = y2
y = ex, y’ = ex,
slope of normal =
By hit and trial we get
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