Question

Divide 70 in to parts so that .Their product is maximum and sum of their square is minimum

Answer 1

If we divide 70 into to parts so we get 23

Answer 2

Answer:  The two parts are 35 and 35.

Step-by-step explanation:

Let the one part is x,

Thus, the second part  = 70 - x

Since, their product = x(70-x)

Let, f(x) = x(70-x)

⇒ f'(x) = -x + 70 - x = -2x + 70  ( By differentiating f(x) with respect to x )

For, maximum or minimum,

f'(x) = 0

⇒ -2x + 70 = 0 ⇒ x = 35

f''(x) = - 2   ( By differentiating f'(x) with respect to x ),

Since, at x = 35, f''(x) = Negative,

Hence, for the value of x = 35 f(x) is maximum,

Now, The sum of the square of x and (70-x) is,

x² + (70-x)²

Let, h(x) = x² + (70-x)²

⇒ h'(x) = 2x - 2 (70 - x)    ( By differentiating h(x) with respect to x )

For maximum or minimum,

h'(x) = 0

⇒ 2x - 2 (70 - x) = 0 ⇒ 4x - 140 = 0 ⇒ x = 35  

h''(x) = 2 - 2 (-1) = 4  ( By differentiating h'(x) with respect to x )

Since, for x = 35, h''(x) = positive

Hence, for x = 35, h(x) is minimum,

Thus, for x = 35, f(x) is maximum and h(x) is minimum.

Thus, the required part of 70, for which their product is maximum and the sum of their square is minimum are, 35 and 35.