Real Analysis 559showing thatg 1 is continuous. This concludes the solution.
502.First, observe that if|x|+|y|→∞thenf (x, y)→∞, hence the function indeed
has a global minimum. The critical points offare solutions to the system of equations
∂f
∂x(x, y)= 4 x^3 + 12 xy^2 −9
4
= 0 ,
∂f
∂y(x, y)= 12 x^2 y+ 4 y^3 −7
4
= 0.
If we divide the two equations by 4 and then add, respectively, subtract them, we obtain
x^3 + 3 x^2 y+ 3 xy^2 +y^3 − 1 =0 andx^3 − 3 x^2 y+ 3 xy^3 −y^3 =^18. Recognizing the perfect
cubes, we write these as(x+y)^3 =1 and(x−y)^3 =^18 , from which we obtainx+y= 1
andx−y=^12. We find a unique critical pointx =^34 ,y=^14. The minimum offis
attained at this point, and it is equal tof(^34 ,^14 )=−^5132.
(R. Gelca)
503.The diameter of the sphere is the segment that realizes the minimal distance between
the lines. So ifP(t+ 1 , 2 t+ 4 ,− 3 t+ 5 )andQ( 4 s− 12 ,−t+ 8 ,t+ 17 ), we have to
minimize the function|PQ|^2 =(s− 4 t+ 13 )^2 +( 2 s+t− 4 )^2 +(− 3 s−t− 12 )^2
= 14 s^2 + 2 st+ 18 t^2 + 82 s− 88 t+ 329.To minimize this function we set its partial derivatives equal to zero:28 s+ 2 t+ 82 = 0 ,
2 s+ 36 t− 88 = 0.This system has the solutiont=− 782 /251,s= 657 /251. Substituting into the equation
of the line, we deduce that the two endpoints of the diameter areP(−^531251 ,−^560251 ,^3601251 )and
Q(−^384251 ,^1351251 ,^4924251 ). The center of the sphere is 5021 (− 915 , 791 , 8252 ), and the radius
√^147
1004. The equation of the sphere is
( 502 x+ 915 )^2 +( 502 y− 791 )^2 +( 502 z− 8525 )^2 = 251 ( 147 )^2.(20th W.L. Putnam Competition, 1959)
504.WritingC=π−A−B, the expression can be viewed as a function in the independent
variablesAandB, namely,f(A,B)=cosA+cosB−cos(A+B).And becauseAandBare angles of a triangle, they are constrained to the domainA, B >0,
A+B<π. We extend the function to the boundary of the domain, then study its extrema.
The critical points satisfy the system of equations