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I have done much research on this question but cannot seem to find any similar to it. The wording does not make sense to me.

Find the projection P onto each of the coordinate planes. Find how the area of P relates to the area of the projections P(xy), P(yz), and P(zx).

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    Presumably $P$ is some subset of $\mathbb{R}^3$. So what other information about $P$ is given? And why aren't we talking about the volume of $P$ rather than the area of $P$? Ok, perhaps the question is about the surface area of $P$. Yes? – 2017-01-26
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    Ok, so the projections are effectively $2$-dimensional "snapshots" of $P$, each from a different perspective. So for example, if $P(xy), P(yz), P(zx)$ have areas $2,3,4$ respectively (just making up some numbers), what can be said about the surface area of $P$? Would that be a particular case of your question? – 2017-01-26
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    For the example I gave above, an obvious lower bound on the surface area of $P$ is $2\times 4 = 8$. An intuitive upper bound is the area of a rectangular $2 \times 3 \times 4$ box, which is $2 \times (2^2 + 3^2 + 4^2) = 58$. – 2017-01-26
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    Oops -- my upper bound above is wrong -- will correct. – 2017-01-26
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    There is no other information given, by I believe that your assumption about it being about surface is correct. This is the major problem I have maybe I just need an equation? – 2017-01-26
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    Yes, I think you need an equation of the surface $P$. Is this a problem from a book? – 2017-01-26
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    It is from a series of questions my professor came up with himself. – 2017-01-26
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    So it's a vague, open-ended question. What is the course? – 2017-01-26
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    It is for Calculus 3. – 2017-01-26
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    Correcting my earlier error, for the case where $P(xy), P(yz), P(zx)$ have areas $2,3,4$, respectively, an intuitive upper bound is $2 \times (2 + 3 + 4) = 18$. – 2017-01-26
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    Was the question intended just as an exploration? Is it a required submission, or just an optional challenge? – 2017-01-26
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    It's required but he likes for the questions to be difficult, requiring us to think. – 2017-01-26
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    It's clearly open ended, with no definite answer. My recommendation is work out the 4 areas (the surface area and the 3 projected areas) for some simple cases -- rectangular solid, sphere, ellipsoid, a very simple tetrahedron (for example the tetrahedron in the first orthant between the coordinate planes and the plane whose equation is $x + y + z = 1$). Submit a few of those. Try to find two examples $P_1$ and $P_2$ for which the projected areas are the same, but the surface areas of $P_1$ and $P_2$ are not equal, thus showing that the projected areas don't determine the surface area. – 2017-01-26
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    Okay thank you so much – 2017-01-26

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