by Guest » Tue Jul 09, 2019 3:44 pm
The volume of a pyramid with square base of side length x and height h is [tex]\frac{x^2h}{3}[/tex]. The surface consists of four triangles and the square base. Obviously the square base has area [tex]x^2[/tex]. The area of each triangular face is the base length, x, times the altitude of each triangle measured along the perpendicular from the vertex of the pyramid perpendicular to the base. Drop a line from the vertex of the pyramid to the center of the base of the pyramid, then the line from there to the center of a side of the base, then a the line back to the vertex. That describes a right triangle with legs of length h and x/2. The altitude of a triangular side is the hypotenuse of that right triangle so has length [tex]\sqrt{h^2+ \frac{x^2}{4}}= 2\sqrt{4h^2+ x^2}[/tex]. The area of each such triangle is "1/2 base times height" so is [tex]\frac{1}{2}x(2\sqrt{4h^2+ x^2}= x\sqrt{4h^2+ x^2}[/tex]. There are 4 such triangles so the total surface area of the pyramid, including all such triangles and the base is [tex]x^2+ 4x\sqrt{4h^2+ x^2}[/tex].
So the problem is to find the value of x that, for fixed h, minimizes [tex]V= x^2+ 4x\sqrt{4h^2+ x^2}[/tex] subject to the constraint [tex]A= \frac{x^2h}{3}= 1000[/tex].