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#?

2 Answers
Jan 16, 2018

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

Jan 16, 2018

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)#