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meerkat18
This problem is a combination of algebra and calculus. First, we formulate an equation for the product. Let x be the first positive number and y be the other positive number. So, the equation would be

xy = 192

Let's rearrange this to make the equation explicit: y = 192/x
Then, the other equation is for the sum. Let S be the sum of the two positive numbers.

S = x + y

Now, substitute the other equation into this one:

S = x + 192/x

Here's where we apply calculus. We can determine the minimum S by getting the first derivative of S with respect to y and equating to zero. Thus,

dS/dx = 1 - 192x⁻² = 0
1 = 192x⁻²
x² = 192
x = +/- √192

But since we are finding the positive number,  x = + √192
Then, we use this to the first equation:

y = 192/√192 = √192

Therefore, the two positive numbers are equal which is √192 or 13.86. 
facundo3141592

The two numbers with a product equal to 192 such that their sum is minimized are:

A = √192 and B = √192

How to find the two numbers?

Let's define A and B as our two numbers, we know that their product must be equal to 192, then we have:

A*B = 192.

Now, the sum of these two numbers is:

A + B.

And we want to minimize this, to do it, we need to use the first equation to rewrite one variable in terms of the other. For example, if we isolate A, we get:

A = 192/B

Replacing this in the sum, we get:

192/B + B

To minimize this, we need to find the values of B that make 0 the differentiation of the above expression.

The differentiation is:

-192/B^2 + 1

Then we need to solve:

-192/B^2 + 1 = 0

192/B^2 = 1

192 = B^2

√192 = B

To get the value of A, we use:

A = 192/B = 192/√192  = √192

Then we can conclude that the two positive numbers such that their product is 192, and their sum is minimized, is:

A = √192 and B = √192.

If you want to learn more about minimization, you can read:

https://brainly.com/question/18585083

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