Let $x$ and $y$ be positive real numbers such that $2x + 3y = 5.$ Find the minimum value of \[\frac{3}{x} + \frac{2}{y}.\]
+2862 https://web2.0calc.com/questions/instant-latex-conversion-to-handle-those-lazy-guests
For the other users that see this problem, the link above will save you time. You can try it out on this problem!
+2862 \(\frac{3}{x} + \frac{2}{y}\)
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)
\(\boxed{5}\)
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
+2862 You have to make sure it satisfies 2x + 3y = 5
How can you be sure xy is equal to 5?
