Maximum value of a variable from a system of 2 equations

Question

From the following $2$ equations, find the maximum value of $d$.
$a + b + c + d = 8$ and $ab + ac + ad + bc + bd + cd = 12$
How to go about with this problem?
Please help.
Thankyou.

Answer 1

Without loss of generality we can assume $a = p + q$, $b = p - q$, with $q > 0$.

The first equation is simplified as

(1) $$ c + d = 8 - 2 \, p. $$

Similarly, for the second equation, we have \begin{align} 12 & = c d + (a + b)(c+ d) + a b \\ & = c d + 2\,p (c + d) + p^2 - q^2, \\ & = c d + 2\,p (8 - 2p) + p^2 - q^2 \\ & = c d + 16\,p - 3\,p^2 - q^2, \end{align} where we have used (1) on the third line.

Thus, \begin{align} c + d &= 8 - 2 \, p, \\ cd &= 12 -16 \, p + 3 \, p^2 + q^2 \end{align} This means that $c$ and $d$ are the roots of the quadratic equation, $$ x^2 - (8 - 2 \, p) \, x + 12 - 16 \, p + 3 \, p^2 +q^2 = 0. $$ In other words, \begin{align} c &= 4 - p - \sqrt{4 + 16 \, p - 2 \, p^2 - q^2} \\ d &= 4 - p + \sqrt{4 + 16 \, p - 2 \, p^2 - q^2}, \end{align} where we have assumed that $d$ gets the plus sign, because we want to maximize $d$.

Now the expression is $d$ obviously maximized with $q = 0$, so let us assume that, \begin{align} d &= 4 - p + \sqrt{4 + 16 \, p - 2 \, p^2} \\ &= (4 - p) + \sqrt{36 - 2 \, (4-p)^2} \end{align}

Finally, let us ask for what value of $p$, $d$ is maximized? For this, let us replace $4 - p = 3 \, \sqrt{2} \, \sin t$, then we have \begin{align} d &= 3 \sqrt{2} \, \left( \sin t + \sqrt{2} \cos t \right) \\ &= 3 \sqrt{2} \, \sqrt{3} \, \left(\sqrt{\frac 1 3} \sin t + \sqrt{\frac 2 3} \cos t \right) \\ &= 3 \sqrt{6} \, \sin\left(t + \tan^{-1}\sqrt2 \right) \\ &\le 3 \sqrt{6}. \end{align}