Mystical box

We must construct the "hyperrectangular polynomial" of this box. Let the box have dimensions $x$ , $y$ , and $z$ . By inspection, we know that: $4(x+y+z)=28\Rightarrow x+y+z=7$ Now the surface area of the box is $2(xy+xz+yz)$ and the volume is $xyz$ . This piques our interest. It seems that the Lebesgue measures (perimeter, area, volume, etc.) of the box are all expressions of the symmetric sums of the three dimensions. This is the motivation behind the hyperrectangular polynomial. For a 3-orthotope (rectangular prism), it is a cubic polynomial whose roots are the dimensions of the prism, and whose coefficients can be used to find the Lebesgue measures of the prism. So the three roots of the hyperrectangular polynomial are $(x,y,z)$ and we have already derived the sum to be 7, hence by Vieta's Formulae and earlier deductions, we note that the polynomial is of the form: $d^3-7d^2+\frac{A}{2}d-V$ Where $A$ is the surface area and $V$ is the volume. Note that we are also given $5A=6V$ , so we can substitute: $d^3-7d^2+\frac{3V}{5}d-V$ If we find the remaining two symmetric sums of the dimensions, we can find the volume and the surface area. We know the sum of our dimensions when taken to different powers. We could use a great deal of algebra to find the remaining symmetric sums, or we can utilize Newton's Sums for the sum of the cubes of the roots to get: $(x^3+y^3+z^3)-7(x^2+y^2+z^2)+\frac{3V}{5}(x+y+z)-3V=0$ Making all known substitutions: $103-7(19)+\frac{3V}{5}(7)-3V=0\Rightarrow V=25$ So the volume of the box is 25. By our ratio, it follows that the surface area is 30. Now that we have found all Lebesgue measures, we can write our hyperrectangular polynomial as: $d^3-7d^2+15d-25$ By the Rational Root Theorem, Synthetic Division, and the Quadratic Formula, we find the roots to be: $d\in\{5,1\pm2i,1\mp2i\}$ But by definition, the roots of the the hyperrectangular polynomial must be the dimensions of the box. However, two of the roots are nonreal, hence the box cannot exist. Q.E.D.

Alternatively, note that after the volume is found to be 25, we can complete the proof without solving for the surface area (and the dimensions). We have: $V=xyz=25$ And: $x+y+z=7$ So we have: $\frac{x+y+z}{3}<\sqrt[3]{xyz}$ However, this contradicts the AM-GM Inequality. Hence, the dimensions are not all nonnegative real numbers so our box cannot exist. Q.E.D.