Minimum of $xy+yz+xz=k$

Question

Given that the sum of x,y,z is 3 find the minimum of xy from the relation$$xy+yz+xz=k$$ Is there anything wrong with my solution since someone said the correct answer differs? $$xy=k-z(x+y)<=>xy=k-z(3-z)=k+z^2-3z=k+\left(z-\frac{3}{2}\right)^2-\frac{9}{4}$$ So $xy\ge k-\frac{9}{4}$ Can anyone give a different solution to confirm?

Answer 1

With Lagrange multipliers the problem can be formulated as

$$ L(x,y,z,\lambda_1,\lambda_2) = x y + \lambda_1(x+y+z-3)+\lambda_2(xy+yz+xz-k) $$

The stationary points are the solutions for

$$ \lambda_1 + y + \lambda_2 (y + z) = 0\\ \lambda_1 + x + \lambda_2 (x + z) = 0\\ \lambda_1 + \lambda_2 (x + y) = 0\\ x + y + z = 3\\ x y + x z + y z = k $$

The solution gives

$$ x y = k - \frac{9}{4} $$

NOTE

$$ x y = \frac{1}{9}(3\pm \sqrt{9-3k})^2 $$

are solutions also but $k - \frac{9}{4}\le \frac{1}{9}(3\pm \sqrt{9-3k})^2 $

The full set of solutions is

$$ \begin{array}{ccccc} x & y & z & \lambda_1 & \lambda_2\\ \frac{1}{4} \left(3-\sqrt{45-16 k}\right)&\frac{1}{4} \left(\sqrt{45-16 k}+3\right)& \frac{3}{2}& \frac{3}{2}& -1\\ \frac{1}{4} \left(\sqrt{45-16 k}+3\right)& \frac{1}{4} \left(3-\sqrt{45-16k}\right)& \frac{3}{2}& \frac{3}{2}& -1\\ \sqrt{1-\frac{k}{3}}+1&\sqrt{1-\frac{k}{3}}+1& 1-2 \sqrt{1-\frac{k}{3}}& -\frac{2 \left(\left(\sqrt{9-3 k}+6\right) k-6 \left(\sqrt{9-3k}+3\right)\right)}{9 (k-3)}& \frac{k-\sqrt{9-3 k}-3}{3 (k-3)}\\ 1-\sqrt{1-\frac{k}{3}}&1-\sqrt{1-\frac{k}{3}}& 2 \sqrt{1-\frac{k}{3}}+1& \frac{2 \left(\left(\sqrt{9-3 k}-6\right) k-6 \sqrt{9-3 k}+18\right)}{9 (k-3)}& \frac{k+\sqrt{9-3 k}-3}{3 (k-3)} \end{array} $$

The minimum solutions are the first and second in the set.

Answer 2

The solution is wrong when $k=\frac{57}{20}.$

You use condition $(x+y)^2\ge 4xy$ to get the domain of $z$.

Then you find the maximum of $xy= g(z)$ on the domain of $z$.