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2 years ago

Given that xy=3/2 and both x and y are nonnegative real numbers, find the minimum value of 10x (3y/5)

Mathematics

1 answer:

Stells [14]2 years ago

5 0

The munimum value is, at x = 3/10, y = 5 and 10x+3y/5 = 6.

<h3>What are nonnegative real numbers?</h3>

Non-negative real numbers are the set of positive real numbers that are bigger than 0 (zero).
That is, the true values are either positive or negative.
The collection will include numbers such as 0, 1, 2, 3, 4, 5, and so on.

To find the minimum value of 10x (3y/5):

Given - y = 3/(2x)

So, we want the bare minimum of,

= 10x + 3/5 × 3/(2x)
= 10x + 9/(10x)

Take the derivative to obtain:

= 10 - 9/(10x^2)

When you set it to zero, you get:

= 100x^2 - 9 = 0
= (10x-3)(10x+3) = 0
= x = ± 3/10

So, at x = 3/10, y = 5 and 10x + 3y/5 = 3 + 3 = 6.

Therefore, the munimum value is, at x = 3/10, y = 5 and 10x+3y/5 = 6.

Know more about nonnegative real numbers here:

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