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

 Nov 10, 2019
 #1
avatar+1679 
0

Can you please un-latex it?

 Nov 10, 2019
 #2
avatar+2490 
+1

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!

CalculatorUser  Nov 10, 2019
 #3
avatar+2490 
+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)

 

 

\(\boxed{5}\)

.
 Nov 10, 2019
edited by CalculatorUser  Nov 10, 2019
 #4
avatar+141 
+1

*Late drumroll*

MagicKitten  Nov 10, 2019
 #8
avatar
+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
avatar+2490 
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
 #10
avatar
0

3/x + 2/y = (3y + 2x)/xy, so wrong at line 3

 

Correct answer is x = 5/4  y = 5/6, giving a minimum of 4.8.

Guest Nov 10, 2019
 #5
avatar
+1

CU: If x = 2.5 and y = 2.5, then how do you balance his equation: 2x +3y =5 ???!!!

 Nov 10, 2019
 #6
avatar+2490 
0

OOPS

CalculatorUser  Nov 10, 2019
 #7
avatar+2490 
+1

fixed it!

CalculatorUser  Nov 10, 2019

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