jlo2ni5x

Answered

2022-07-21

Finding value of t such that volume contained inside the planes is minimum

The question is to find out the value of t such that volume contained inside the planes $\sqrt{1-{t}^{2}}x+tz=2\sqrt{1-{t}^{2}},$

$z=0,$

$x=2+\frac{t\sqrt{4{t}^{2}-5t+2}}{\sqrt{12}(1-{t}^{2}{)}^{1/4}}\text{and}$

$|y|=2$ 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.

The question is to find out the value of t such that volume contained inside the planes $\sqrt{1-{t}^{2}}x+tz=2\sqrt{1-{t}^{2}},$

$z=0,$

$x=2+\frac{t\sqrt{4{t}^{2}-5t+2}}{\sqrt{12}(1-{t}^{2}{)}^{1/4}}\text{and}$

$|y|=2$ 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.

Answer & Explanation

Abbigail Vaughn

Expert

2022-07-22Added 15 answers

Step 1

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 $(x,z)=(c,-\frac{a(c-2)}{b})$ respectively. Thus the base is a right triangle with catheti $|c-2|$ and $|\frac{a(c-2)}{b}|$.

Step 2

Resulting volume is $V=\frac{1}{2}\ast |\frac{t\sqrt{4{t}^{2}-5t+2}}{\sqrt{12}(1-{t}^{2}{)}^{1/4}}|\ast 4\ast |{z}_{0}|$ where ${z}_{0}$ is where planes (1) and (3) intersect.

$2-\frac{t}{\sqrt{1-{t}^{2}}}{z}_{0}=2+\frac{t\sqrt{4{t}^{2}-5t+2}}{\sqrt{12}(1-{t}^{2}{)}^{1/4}}\phantom{\rule{0ex}{0ex}}{z}_{0}=-\frac{1}{\sqrt{12}}(1-{t}^{2}{)}^{1/4}\sqrt{4{t}^{2}-5t+2}\phantom{\rule{0ex}{0ex}}V=\frac{1}{6}|t|(4{t}^{2}-5t+2)\phantom{\rule{thinmathspace}{0ex}}|t\in (-1,1)$

But this V(t) has no maximum; its limit as $t\to -1$ is $\frac{11}{6}$, greater than any value on given interval.

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 $(x,z)=(c,-\frac{a(c-2)}{b})$ respectively. Thus the base is a right triangle with catheti $|c-2|$ and $|\frac{a(c-2)}{b}|$.

Step 2

Resulting volume is $V=\frac{1}{2}\ast |\frac{t\sqrt{4{t}^{2}-5t+2}}{\sqrt{12}(1-{t}^{2}{)}^{1/4}}|\ast 4\ast |{z}_{0}|$ where ${z}_{0}$ is where planes (1) and (3) intersect.

$2-\frac{t}{\sqrt{1-{t}^{2}}}{z}_{0}=2+\frac{t\sqrt{4{t}^{2}-5t+2}}{\sqrt{12}(1-{t}^{2}{)}^{1/4}}\phantom{\rule{0ex}{0ex}}{z}_{0}=-\frac{1}{\sqrt{12}}(1-{t}^{2}{)}^{1/4}\sqrt{4{t}^{2}-5t+2}\phantom{\rule{0ex}{0ex}}V=\frac{1}{6}|t|(4{t}^{2}-5t+2)\phantom{\rule{thinmathspace}{0ex}}|t\in (-1,1)$

But this V(t) has no maximum; its limit as $t\to -1$ is $\frac{11}{6}$, greater than any value on given interval.

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