Problem:
Given that β4β€xβ€β2 and 2β€yβ€4, what is the largest possible value of xx+yβ?
Answer Choices:
A. β1
B. β21β
C. 0
D. 21β
E. 1
Solution:
Because
xx+yβ=1+xyβ and xyβ<0
the value is maximized when β£y/xβ£ is minimized, that is, when β£yβ£ is minimized and β£xβ£ is maximized. So y=2 and x=β4 gives the largest value, which is 1+(β1/2)=21β.
Answer: Dβ.
The problems on this page are the property of the MAA's American Mathematics Competitions