Assuming the pyramid must be a familiar sq. pyramid, the biggest sq. pyramid interior a cube with area length d, could have top f = d and base area e = d (such that it shares a cube's area as its base), and its apex on the middle of the different cube area, such that its quantity V[pyramid] = (a million/3) e^2 f = d^3 / 3 = one-third the quantity V[cube] of the cube. subsequently, V[pyramid] is maximized as V[cube] is maximized. A cube interior a cylinder with radius b and top c, could have quantity V[cube] such that: V[cube] = min(2b, c)^3 *** Eq. a million on the grounds that c is inversely proportional Assuming the two caps of the cylinder are such that their around perimeters are parallel "small circles" of the sphere, (it extremely is, for the cylinders optimum radius b[max] = a million, c = 0; and additionally, for the cylinder's optimum top c[max] = 2, b = 0). enable element o be the middle of the sphere, and enable element p be on the edge of a around cap. enable oq be a radius of the cylinder such that pq is perpendicular to the around cap. Then, op = the around radius a = a million, pq = a million/2 the cylinder's top = c/2, and oq = the cylinder's radius b opq is a suitable triangle with hypotenuse op. via the Pythagorean Theorem, c^2 / 4 + b^2 = a million => c = 2 sqrt(a million - b^2) *** Eq. 2 on the grounds that 0 < b < a million and 0 < c < 2, we want in basic terms evaluate the beneficial branches of the sqrts. Now, b is inversely proportional to c, and the two are monotonic, because of the fact of this (via Eq. a million), V[cube] is maximized the place b = c. So our Pythagorean equation could be simplified to: b^2 / 4 + b^2 = a million => b = 2 sqrt(5) / 5 And, so V[pyramid, max] = V[cube, max] / 3 = b^3 / 3 = (2 sqrt(5) / 5)^3 / 3 = 8 sqrt(5) / seventy 5 *** answer ? 5.7% of the quantity of the sphere --- because of the fact the inradius of a unit cube is a million/2 and a circumradius is sqrt(3)/2, i could think of packing the sphere interior the cube could maximize area utilization. the main awkward quantity to fill seems to be the pyramid. So i could wager a maximal order could bypass like this (from innermost to outermost): pyramid, cylinder, cube, sphere i could wager that a minimum order could be cylinder, cube, sphere, pyramid