Let $x$ and $y$ be positive real numbers such that $2x + 3y = 5.$ Find the minimum value of \[\frac{3}{x} + \frac{2}{y}.\]

Guest Nov 10, 2019

#1

#2**+1 **

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CalculatorUser
Nov 10, 2019

#3**+2 **

\(\frac{3}{x} + \frac{2}{y}\)

Simplifies to

\(\frac{2x+3y}{xy}\)

Then we substitute in \(2x + 3y = 5\)

SO we have, \(\frac{5}{xy}\)

Since x and y MUST be positive real numbers, the product of XY must be as large as possible.

This means:

1) X and Y must be as close to possible

2) X and Y must satisfy 2x + 3y = 5.

3) they are positive

This is easy, the values of X and Y respectively must be (1, 1). They are as close as possible, AND they satisfy the equation

So the minimum value is

\(\frac{5}{1}\)

which is equal to (drumroll please)

\(\boxed{5}\)

.CalculatorUser Nov 10, 2019

#8**+1 **

great solution! Exactly what I did too :))

But apart from "must satisfy the equation 2x+3y=5"

we already found that 5/xy=?? (Minimum value)

therefore, 5=xy :)

so 5/5=1

Guest Nov 10, 2019

#9**0 **

You have to make sure it satisfies 2x + 3y = 5

How can you be sure xy is equal to 5?

CalculatorUser
Nov 10, 2019