Ask your own question, for FREE!
Mathematics 13 Online
OpenStudy (anonymous):

if x,y,z be three non-negative integers such that x+y+z=10. the maximum possible value of xyz+xy+yz+zx is a. 52 b. 64 c. 69 d. 73

OpenStudy (anonymous):

xyz+xy+xz+yz can be expressed as (x + 1)(y + 1)(z + 1) - (x + y + z + 1) Since x+y+z=10 xyz+xy+xz+yz = (x + 1)(y + 1)(z + 1) - 11 So now we have to maximize (x + 1)(y + 1)(z + 1) - 11 Now x+y+z=10 can be written as (x+1) + (y+1)+ (z+1)=13 We can now use AM-GM Rule to get max value, which we will get when value of x+1,y+1 & z+1 is as close as possible (if not equal) Now floor(13/3) = 4 so we can have max value at (4,4,(13-(4+4)) i.e. 4,4,5 So Max value = 4*4*5 - 11 = 69

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!