Question

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

Answer 1

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

What are nonnegative real numbers?

• 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:

brainly.com/question/26606859

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