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The question is to find out the value of $t$ such that volume contained inside the planes $$\sqrt{1-t^2}x+tz=2\sqrt{1-t^2},$$$$z=0,$$$$x=2+\frac{t\sqrt{4t^2-5t+2}}{\sqrt{12}(1-t^2)^{1/4}} \text{ and }$$ $$|y|=2$$ is maximum.

I tried to figure out the line of the intersection of the planes but it made the problem more complicated.Is there any way by which the equations can be decomposed to make it easier to handle?Any help shall be highly appreciated.Thanks.

Navin
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1 Answers1

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Note that there is only one equation on $y$ which means that for our body $y \in [-2,2]$. Three other equations are equations of some planes parallel to $Oy$ axis. Thus the body in question is a right prism with triangle base and height $4$.
Equations (2) and (3) are of orthogonal planes, they intersect at $(x,z) = (c,0)$. Plane (1) is $a(x-2)+bz=0$ and intersects other two at $(x,z) = (2,0)$ and $(x,z) = (c,-{a(c-2) \over b})$ respectively. Thus the base is a right triangle with catheti $|c-2|$ and $|{a(c-2) \over b}|$.

Resulting volume is $V = {1 \over 2}*|{t \sqrt{4t^2-5t+2}\over \sqrt{12} (1-t^2)^{1/4}}|*4*|z_0|$ where $z_0$ is where planes (1) and (3) intersect. $$2 - {t \over \sqrt{1-t^2}}z_0 = 2+{t \sqrt{4t^2-5t+2}\over \sqrt{12} (1-t^2)^{1/4}} \\ z_0 = -{1 \over \sqrt{12}}(1-t^2)^{1/4}\sqrt{4t^2-5t+2} \\ V = {1 \over 6}|t|(4t^2-5t+2) \, | t \in (-1,1) $$ But this $V(t)$ has no maximum; its limit as $t \to -1$ is ${11 \over 6}$, greater than any value on given interval.

Abstraction
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  • Can you please explain how you found out the resulting volume – Navin Mar 02 '17 at 05:01
  • @navinstudent The only restriction on $y$ is $y \in [-2,2]$, so the resulting body is a prism of height $4$. Its base is formed by lines $z=0, x = c, a(x-2)+bz=0$ making it right triangle with one vertex in $x=2, z=0$ and area ${1 \over 2}|c-2||{a(c-2) \over b}|$. – Abstraction Mar 02 '17 at 09:12
  • Can you please explain in more details.I am afraid I couldnot get what do $c,a$and $b$ stand for and how they form the base of the prism so obtained? – Navin Mar 02 '17 at 16:09
  • @navinstudent Added expanded explanation. $a, b, c$ stand for $\sqrt{1-t^2}, t, 2+{t \sqrt{4t^2-5t+2}\over \sqrt{12} (1-t^2)^{1/4}}$ respectively (too long to write repeatedly, especially the last one). – Abstraction Mar 03 '17 at 09:59
  • Thanks for your explanation.I have now understood very well. – Navin Mar 03 '17 at 13:33
  • One slight thing.if we take $f(t)=4t^3-5t^2+2t$ $f'(t)=12t^2-10t+2$ evaluating this to zero we get $t=1/2$ or $t=1/3$ Checking $f''(t)=24t-10$.We can infer that volume is maximum at $t=1/3$.Thanks. – Navin Mar 03 '17 at 13:35
  • @navinstudent Yes, you're right, but you stop halfway. First, we actually have $f(t) = |4t^3-5t^2+2t|$ so there is another minimum at $t=0$. Second, $f(1/3) = 7/27$ while $f(-1) = 11$ so while $t=1/3$ is indeed local maximum, it is not the greatest value our function can take. – Abstraction Mar 03 '17 at 13:47
  • Oh yes I missed out that.Thanks for explanation. – Navin Mar 03 '17 at 13:55