Bradley completed 3/5 of his homework. What percent of his homework does he still need to complete?

Mathematics

1 answer:

svp [43]3 years ago

7 0

40 percent. Times both by twenty

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Find the maximum volume of a rectangular box that is inscribed in a sphere of radius r.

zvonat [6]

Answer:

The maximum volume of a box inscribed in a sphere of radius r is a cube with volume $\frac{8r^3}{3\sqrt{3}}$ .

Step-by-step explanation:

This is an optimization problem; that means that given the constraints on the problem, the answer must be found without assuming any shape of the box. That feat is made through the power of derivatives, in which all possible shapes are analyzed in its equation and the biggest -or smallest, given the case- answer is obtained. Now, 'common sense' tells us that the shape that can contain more volume is a symmetrical one, that is, a cube. In this case common sense is correct, and the assumption can save lots of calculations, however, mathematics has also shown us that sometimes 'common sense' fails us and the answer can be quite unintuitive. Therefore, it is best not to assume any shape, and that's how it will be solved here.

The first step of solving a mathematics problem (after understanding the problem, of course) is to write down the known information and variables, and make a picture if possible.

The equation of a sphere of radius $r$ is $x^2 + y^2 + z^2=r^2$ . Where $x$ , $y$ and $z$ are the distances from the center of the sphere to any of its points in the border. Notice that this is the three-dimensional version of Pythagoras' theorem, and it means that a sphere is the collection of coordinates in which the equation holds for a given radius, and that you can treat this spherical problem in cartesian coordinates.

A box that touches its corners with the sphere with arbitrary side lenghts is drawn, and the distances from the center of the sphere -which is also the center of the box- to each cartesian axis are named $x$ , $y$ and $z$ ; then, the complete sides of the box are measured $2x$ , $2y$ and $2z$ . The volume $V$ of any rectangular box is given by the product of its sides, that is, $V=2x\cdot 2y\cdot 2z=8xyz$ .

Those are the two equations that bound the problem. The idea is to optimize $V$ in terms of $r$ , therefore the radius of the sphere must be introduced into the equation of the volumen of the box so that both variables are correlated. From the equation of the sphere one of the variables is isolated: $z^2=r^2-x^2 - y^2\quad \Rightarrow z= \sqrt{r^2-x^2 - y^2}$ , so it can be replaced into the other: $V=8xy\sqrt{r^2-x^2 - y^2}$ .

But there are still two coordinate variables that are not fixed and cannot be replaced or assumed. This is the point in which optimization kicks in through derivatives. In this case, we have a cube in which every cartesian coordinate is independent from each other, so a partial derivative is applied to each coordinate independently, and then the answer from both coordiantes is merged into a single equation and it will hopefully solve the problem.

The $x$ coordinate is treated first: $\frac{\partial V}{\partial x} =\frac{\partial 8xy\sqrt{r^2-x^2 - y^2}}{\partial x}$ , in a partial derivative the other variable(s) is(are) treated as constant(s), therefore the product rule is applied: $\frac{\partial V}{\partial x} = 8y\sqrt{r^2-x^2 - y^2} + 8xy \frac{(r^2-x^2 - y^2)^{-1/2}}{2} (-2x)$ (careful with the chain rule) and now the expression is reorganized so that a common denominator is found $\frac{\partial V)}{\partial x} = \frac{8y(r^2-x^2 - y^2)}{\sqrt{r^2-x^2 - y^2}} - \frac{8x^2y }{\sqrt{r^2-x^2 - y^2}} = \frac{8y(r^2-2x^2 - y^2)}{\sqrt{r^2-x^2 - y^2}}$ .

Since it cannot be simplified any further it is left like that and it is proceed to optimize the other variable, the coordinate $y$ . The process is symmetrical due to the equivalence of both terms in the volume equation. Thus, $\frac{\partial V}{\partial y} = \frac{8x(r^2-x^2 - 2y^2)}{\sqrt{r^2-x^2 - y^2}}$ .

The final step is to set both partial derivatives equal to zero, and that represents the value for $x$ and $y$ which sets the volume $V$ to its maximum possible value.

$\frac{\partial V}{\partial x} = \frac{8y(r^2-2x^2 - y^2)}{\sqrt{r^2-x^2 - y^2}} =0 \quad\Rightarrow r^2-2x^2 - y^2=0$ so that the non-trivial answer is selected, then $r^2=2x^2+ y^2$ . Similarly, from the other variable it is obtained that $r^2=x^2+2 y^2$ . The last equation is multiplied by two and then it is substracted from the first, $r^2=3 y^2\therefore y=\frac{r}{\sqrt{3}}$ . Similarly, $x=\frac{r}{\sqrt{3}}$ .

Steps must be retraced to the volume equation $V=8xy\sqrt{r^2-x^2 - y^2}=8\frac{r}{\sqrt{3}}\frac{r}{\sqrt{3}}\sqrt{r^2-\left(\frac{r}{\sqrt{3}}\right)^2 - \left(\frac{r}{\sqrt{3}}\right)^2}=8\frac{r^2}{3}\sqrt{r^2-\frac{r^2}{3} - \frac{r^2}{3}} =8\frac{r^2}{3}\sqrt{\frac{r^2}{3}}=8\frac{r^3}{3\sqrt{3}}$ .