let a,b,c be non negative integers. a+b+c=10 what is the max value of abc+ab+bc+ca

f(a,b,c)=ab+c(a+b)

=ab+(10-(a+b))(a+b)

=ab+10(a+b)-(a+b)^2

Thus, f(a,b)= ab+10(a+b)-(a+b)^2

p=b+10–2(a+b) …(1) [first derivative of f()wrt a]

q=a+10–2(a+b) …(2) [first derivative of f()wrt b]

Solving 1 and 2, by equating p=0,q=0, we get,

a=b=10/3 (these are stationary points)

r=-2 [2nd derivative of f() wrt a]

t=-2 [2nd derivative of f() wrt b]

s=-1 [1st derivative of (1) wrt b] or [1st derivative of (2) wrt a]

Now,

rt-s^2= 3>0, and r

Hence, for f(a,b) to be maximum, a=10/3, b=10/3. Then c=10-(a+b)=10–(20/3)=10/3

Thus, fmax= (ab+bc+ca) |a=b=c=10/3

= 300/9= 33.3333… (Ans)