How do you solve this question? When 1<=x<=81x8 and log_2(y)=[log_2(x)]^2log2(y)=[log2(x)]2 is given, What is maximum and minimum value of x^2/yx2y?

2 Answers
Jan 16, 2018

Maximum value = 2

Explanation:

When we put xx = 2 in the equation
log_2(y) = [log_2(x)]^2log2(y)=[log2(x)]2
We get,
log_2(y) = [log_2(2)]^2log2(y)=[log2(2)]2
log_2(y) = (1)^2log2(y)=(1)2
log_2(y) = 1log2(y)=1
We get y = 2
Put the values of x and y in the equation x^2/yx2y
= 2^2/2222
= 2
This is the maximum value
I am not sure for the minimum
It should be 1/818 as per my concern by putting x = 8

Hope it helps you

Jan 16, 2018

See below.

Explanation:

From log_2y = (log_2 x)^2log2y=(log2x)2

making x = 2^kx=2k we have

log_2 y = k^2 rArr y = 2^(k^2)log2y=k2y=2k2

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)