Find two positive numbers satisfying the given requirements. The sum of the first and twice the second is 80 and the product is a maximum.

Solution:

Let the first number be x and the second number be y.

Given, x + 2y = 80

⇒ 2y = 80 - x

⇒ y= 40 - x/2

Product, M = xy

Product of two numbers is maximum if their derivative is zero i.e. f(x) = xy = 0.

Put value of y in M = xy

⇒ M = x (40 - x/2)

⇒ M = 40x - x²/2

On differentiating,

dM/dx = 40 - 2x/2

= 40 - x

For product to be maximum, 40 - x = 0

⇒ x = 40

Substitute x = 40 in x + 2y = 80

⇒ 40 + 2y = 80

⇒ 2y = 40

⇒ y = 20

Therefore, the two positive numbers are 40 and 20.

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Find two positive numbers satisfying the given requirements. The sum of the first and twice the second is 80 and the product is a maximum.

Summary:

Two positive numbers such the sum of the first and twice the second is 80 and whose product is maximum is x = 40 and y = 20.