Find $MAX$ and $MIN$ of $P = 2ab+3ac+3bc+\frac{6}{a+b+c}$

Question

Let : $a,b,c\in\mathbb{R}$ , $a\ge 0 , b\ge 0,0\le c\le 1 $ and $a^2+b^2+c^2=3$.

Find $MAX$ and $MIN$ of $P=2ab+3ac+3bc+\dfrac{6}{a+b+c}$.

Answer 1

Apply Lagrange multiplier, but you will have to add 5 conditions, the inequalities will become equalities, so that will give $2^4=32$ cases to check. This is actually application of the KKT conditions Although some of them will cancel, because of contradiction. The function is:

$$f(a,b,c) = 2ab + 3ac + 3bc + \frac{6}{a+b+c}$$

After applying Lagrange Multiplier we have:

$$F(a,b,c,\lambda,\lambda_1, \lambda_2, \lambda_3, \lambda_4) = 2ab + 3ac + 3bc + \frac{6}{a+b+c} - \lambda(a) - \lambda_1(b) - \lambda_2(c) + \lambda_3(c-1) + \lambda_4(a^2 + b^2 + c^2 - 3)$$

Now we take partial derivatives:

$$F_a = 2b + 3c -\frac{6}{(a+b+c)^2} - \lambda + 2\lambda_4a= 0$$ $$F_b = 2a + 3c -\frac{6}{(a+b+c)^2} - \lambda_1 + 2\lambda_4b= 0$$ $$F_c = 3a + 3b -\frac{6}{(a+b+c)^2} - \lambda_2 + \lambda_3 + 2\lambda_4c= 0$$

We know that those $\lambda$ terms has to be set to 0, so we have $16$ distinct cases.

Case 1: $\lambda = \lambda_1 = \lambda_2 = \lambda_3 = 0$

Case 2: $\lambda = \lambda_1 = \lambda_2 = c-1 = 0$

Case 3: $\lambda = \lambda_1 = c = \lambda_3 = 0$

This one is violates the condition: $0<c$

Case 4: $\lambda = \lambda_1 = c = c-1 = 0$

This is a contradiction.

Case 5: $\lambda = b = \lambda_2 = \lambda_3 = 0$

Case 6: $\lambda = b = \lambda_2 = c-1 = 0$

This implies a solution $(a,b,c) = (\sqrt{2},0,1)$

Case 7: $\lambda = b = c = \lambda_3 = 0$

This one is violates the condition: $0<c$

Case 8: $\lambda = b = c = c-1 = 0$

This is a contradiction.

Case 9: $a = \lambda_1 = \lambda_2 = \lambda_3 = 0$

Case 10: $a = \lambda_1 = \lambda_2 = c-1 = 0$

This implies a solution $(a,b,c) = (0,\sqrt{2},1)$

Case 11: $a = \lambda_1 = c = \lambda_3 = 0$

This one is violates the condition: $0<c$

Case 12: $a = \lambda_1 = c = c-1 = 0$

This is a contradiction.

Case 13: $a = b = \lambda_2 = \lambda_3 = 0$

This implies a solution $(a,b,c) = (0,0,\sqrt{3})$, but this violates the condition $c\le1$

Case 14: $a = b = \lambda_2 = c-1 = 0$

This implies a solution $(a,b,c) = (0,0,1)$, this implies the condition $a^2 + b^2 + c^2 = 3$

Case 15: $a = b = c = \lambda_3 = 0$

This one is violates the condition: $0<c$

Case 16: $a = b = c = c-1 = 0$

This is a contradiction.

Now as you can see we excluded many cases, and i solved some that are easier, now you can play with the other $3-4$ cases using the partial derivatives.

Now plug all those solutions into the initial function and see which one is minima and which one is maxima.