is this question complete ??
BTW maximum value of the product of x and y is 45*45/12*14
this is attained at y = 45/14 and x= 45/12
6x + 7y = 45
=> x = (45 - 7y)/6
Let z = x*y
=> z = y*(45 - 7y)/6
Diffrenciate with respect to y and equate it to 0 (for maximum)
solve for y to get y = 45/14
=> x = 45/12 (from 6x + 7y = 45)
therefor x*y = 45*45/12*14
there are some answers wrong in GRE prep CD's
More over 6x + 7y = 45 and xy can be equal but for that x, y has to be complex numbers (if GRE has complex numbers in course then probably D is the answer)
I guess u dont have to actually solve such problems at all...Just plug-in some numbers and cross check....
Case 1 : if x = 0; 6x+7y = 45 (x = 45/6) and xy = 0 ; So, A is greater
Case 2 : if x = -1; 6x+7y = 45 (y = 51/7) and xy = -51/7 ; So, A is greater
Case3 : if x = 1/6; 6x+7y = 45 (y = 44/6) and xy = 44/36 ; So again A is greater
Case 4 : if x = -1/6 again xy is negative and 6x+7y is positive, therefore A is greater...
Since it satisfies all the conditions I would go with A as well...Yah, there are some mistakes in the Peterson's guide!!.
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