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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)
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