# Find two positive numbers x and y such that their sum is 35 and the product x^{2} y^{5} is a maximum

**Solution:**

Maxima and minima are known as the extrema of a function

Maxima and minima are the maximum or the minimum value of a function within the given set of ranges.

Let one number be x.

Then, the other number is y = (35 - x).

Let P (x) = x^{2} y^{5}

Then we have,

P (x) = x^{2} (35 - x)^{5}

Therefore,

P' (x) = 2x (35 - x)^{5} - 5x^{2} (35 - x)^{4}

= x (35 - x)^{4} [2 (35 - x) - 5x]

= x (35 - x)^{4} (70 - 7x)

= 7x (35 - x)^{4} (10 - x)

P" (x) = 7 (35 - x)^{4} (10 - x) + 7x [- (35 - x)^{4} - 4 (35 - x)^{3} (10 - x)]

= 7 (35 - x)^{4} (10 - x) - 7x (35 - x)^{4} - 28x (35 - x)^{3} (10 - x)

= 7 (35 - x)^{3} [(35 - x)(10 - x) - x (35 - x) - 4x (10 - x)]

= 7 (35 - x)^{3} [350 - 45x + x^{2} - 35x + x^{2} - 40x + 4x^{2}]

= 7 (35 - x)^{3} (6x^{2} - 120x + 350)

Now,

P' (x) = 0

⇒ x = 0, x = 35, x = 10

When, x = 35

Then,

P' (x) = P (x) = 0

⇒ y = 35 - 35 = 0

This will make the product x^{2} y^{5} equal to 0.

When, x = 0

Then,

⇒ y = 35 - 0 = 35

This will make the product x^{2} y^{5} equal to 0.

Hence,

x = 0 and x = 35 cannot be the possible values of x.

When, x = 10

Then,

P" (x) = 7 (35 - 10)^{3} (6 x 100 - 120 × 10 + 350)

= 7 (25)^{3} (- 250) < 0

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

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.5 Question 15

## Find two positive numbers x and y such that their sum is 35 and the product x^{2} y^{5} is a maximum

**Summary:**

The two positive numbers x and y such that their sum is 35 and the product x^{2} y^{5} is a maximum are 10 and 25. Maxima and minima are known as the extrema of a function

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