NCERT Solutions
Class 12 Maths
Application of Derivatives

Ex.6.5 Q.15
Find two positive numbers x and y such that their sum is 35 and the product x2y5 is a maximum.
Let one number be x. Then, the other number is y = (35 − x).
Let f(x) = x2y5
⟹ f(x) = x2(35 - x)5
So, f’(x) = 2x (35 - x)5 - 5x2(35 - x)4
⟹ f’(x) = x(35 - x)4[2(35 - x) – 5x]
⟹ f’(x) = x(35 - x)4[70 - 2x – 5x]
⟹ f’(x) = x(35 - x)4(70 - 7x)
Now, f’(x) = 0
⟹ x(35 - x)4(70 - 7x) = 0
⟹ x = 0, 35, 10
Again f”(x) = 7(35 - x)4(10 - x) + 7x[-(35 - x)4 – 4(35 - x)3(10 - x)]
⟹ f”(x) = 7(35 - x)4(10 - x) - 7x (35 - x)4 – 28x(35 - x)3(10 - x)
⟹ f”(x) = 7(35 - x)3[(35 - x)(10 - x) - x(35 - x) – 4x (10 - x)]
⟹ f”(x) = 7(35 - x)3[350 – 45x + x2 – 35x + x2 – 40x + 4x2]
⟹ f”(x) = 7(35 - x)3(6x2 – 120x + 350)
When x = 35, y = 35 – 35 = 0 and the product x2y5 will be equal to 0.
When x = 0, y = 35 − 0 = 35 and the product x2y5 will be 0.
So, x = 0 and x = 35 cannot be the possible values of x.
When x = 10, we have
f”(x) = 7(35 - 10)3(6 (10)2 – 120 (10) + 350)
= 7(25)3(-250)
< 0
So, By second derivative test, P(x) will be the maximum when x = 10 and
y = 35 − 10 = 25
Hence, the required numbers are 10 and 25.