Find two positive numbers that satisfy the following: if xy=20, find the minimum value of 2x+3y
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)`
Therefore when y'=0
`therefore 0=(2x^2-60)/x^2 therefore 0=2x^2-60`
`therefore x=+-sqrt 30`
and we know x is positive
and if we substitute these values into the original expression of 2x+3y to verify the answer:
`therefore 2(sqrt30) + 3(20/(sqrt30))`
`therefore x=sqrt30 and y=20/sqrt30`
or `x=5.477 and y=3.651` (both rounded off)