A rectangular box that rests on the $x, y$ plane is to be inscribed in the cone $z = a − \sqrt{x^2 + y^2}$, $\ a \ge z \ge 0$. Find the dimensions of the box that maximise its volume.
How would I start with this question? My thoughts are to use the cone's equation and find the rectangles vertices. Is this a reasonable assumption and if so, how would I do it?
To construct a box inside the cone you have to choose two things: the height $h$ of the box and it's basis, in other words the rectangle vertices. Then the volume of the box will be $h\cdot A$, where $A$ is the area of the rectangle. Fixed $h$, the basis must be inscribed on the circle obtained sectioning the cone at height $h$. This is because obviously the maximum is obtained when the box is touching the cone on the top. So the problem is to find the rectangle of maximum area inscribed in a circle: it is the square (try to show it). At height $h$ the radius of the circle is $a-h$, so the area of the inscribed square is $$A=2(a-h)^2.$$ So the volume will be $$V=A \cdot h=2h(a-h)^2.$$ We have to maximize $V$ with $0\le h\le a$. To do so we can for example compute derivative $$V'=2[(a-h)^2-2h(a-h)]=2(a-h)(a-3h).$$ So we have the maximum when $h=a/3$, obtaining $V=8a^3/27$. The dimensions of the box will be $a/3\times 2 \sqrt 2 a/3 \times 2 \sqrt 2 a/3.$