Question

How do you solve this question? When `1<=x<=8` and `log_2(y)=[log_2(x)]^2` is given, What is maximum and minimum value of `x^2/y`?

Answer 1

Maximum value = 2

Explanation:

When we put `x` = 2 in the equation
`log_2(y) = [log_2(x)]^2`
We get,
`log_2(y) = [log_2(2)]^2`
`log_2(y) = (1)^2`
`log_2(y) = 1`
We get y = 2
Put the values of x and y in the equation `x^2/y`
= `2^2/2`
= 2
This is the maximum value
I am not sure for the minimum
It should be `1/8` as per my concern by putting x = 8

Hope it helps you

Answer 2

See below.

Explanation:

From `log_2y = (log_2 x)^2`

making `x = 2^k` we have

`log_2 y = k^2 rArr y = 2^(k^2)`

and we have the pairs

`((k, x,y,x^2/y),(1,2,2,1/2),(2,4,16,1/4),(3,8,512,1/8))`

So, for integer values in the feasible region, we have

`min = 1/8` with `(x,y) = (8,512)`
`max = 1/2` with `(x,y) = (2,2)`