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# Help quick

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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
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Nov 10, 2019
#2
+2850
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CalculatorUser  Nov 10, 2019
#3
+2850
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$$\frac{3}{x} + \frac{2}{y}$$

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
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*Late drumroll*

MagicKitten  Nov 10, 2019
#8
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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
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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
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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
+1

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

Nov 10, 2019
#6
+2850
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OOPS

CalculatorUser  Nov 10, 2019
#7
+2850
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fixed it!

CalculatorUser  Nov 10, 2019