# Find two positive numbers that satisfy the following: if xy=20, find the minimum value of 2x+3y

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### 1 Answer

We know that `xy = 20` and therefore we can estimate by using the factors of 20, what our answer will approximate to.

From the factors, 1 x 20, 2 x 10, 4 x 5, we can already see that `x~~ 5` and `y~~ 4` giving us `2(5) + 3(4) = 22` .

Now we use calculus to find the actual answer.

Substitute a value for y such that if `xy=20` ,

`therefore y=20/x` and substituting that into `2x+3y` ,

`therefore = 2x + 3(20/x)` .

Using the rules of exponents:

`therefore = 2x+60(x)^(-1)`

We find the minimum value when y' = 0

`therefore y' = 2 -60 (x)^(-2)`

`= 2-60/x^2`

Therefore when y'=0

`0=2-60/x^2`

`therefore 0=(2x^2-60)/x^2 therefore 0=2x^2-60`

`therefore 2x^2=60`

`therefore x^2=30`

`therefore x=+-sqrt 30`

and we know x is positive

`therefore x=sqrt30`

and `y=20/x`

`therefore y=20/(sqrt30)`

and if we substitute these values into the original expression of 2x+3y to verify the answer:

`therefore 2(sqrt30) + 3(20/(sqrt30))`

= 21.9

**Ans:**

`therefore x=sqrt30 and y=20/sqrt30`

**or** `x=5.477 and y=3.651` **(both rounded off)**