Find the greatest and least values of Volume of cylinder

[chwala]

Homework Statement

The volume of a cylinder is given by the formula $V=πr^2h$. Find the greatest and least values of $V$ if $r+h=6$

Relevant Equations

rate of change- differentiation

$$V=πr^2h$$
$$V=πr^2(6-r)$$
$$\frac {dV}{dr}=12πr-3πr^2$$
For max/min value, $$\frac {dV}{dr}=0$$
$$12πr-3πr^2=0$$
$$3πr(4-r)=0$$
$r=0$ or $r=4$
$$⇒V_{max}= 32π$$
$$⇒V_{min}= 0$$,
I do not think there is another way of doing this...

 

[Mark44]

I do not think there is another way of doing this...

You could find or at least approximate the maximum by inspecting the graph of $V(r) = \pi r^2(6 - r)$. And clearly V(r) = 0 when r = 0. Although there are values of r for which V(r) < 0, since this is a physical cylinder, we wouldn't consider values of r > 6, since those would make the height negative.

 

[Office_Shredder]

Fun fact, you missed a value of r which either minimizes or maximizes the volume of the cylinder. What else do you have to check besides where the derivative is zero?

 

[docnet]

I do not think there is another way of doing this...

The method of Lagrange multipliers is how we learned to solve constrained minimization/maximization problems in class.

 

[docnet]

The goal of the method of Lagrange multipliers is to maximize $f(x,y)$ subject to the constraint $g(x,y)$. Here $f$ and $g$ are assumed to have continuous first partial derivatives. $\lambda$ is a new variable called the Lagrange multiplier. Our goal is to determine the stationary points of the Lagrangian
$$\mathcal{L} (x,y,\lambda)=f(x,y)-\lambda g(x,y)$$
Let $f(x,y)=\pi x^2 y$ and $g(x,y)=x+y-6$. The corresponding Lagrangian is
$$\mathcal{L} (x,y,\lambda)=\pi x^2 y-\lambda (x+y-6)$$
Take partial derivatives as you would compute the gradient of any multi variable function and set it equal to $0$ to find the stationary points.
\begin{align}D_x \mathcal{L}=&2\pi xy-\lambda=0\\
D_y \mathcal{L}=&\pi x^2-\lambda=0\\
g(x,y)=& x+y-6=0\end{align}
Then we look for all the solutions to the system of $3$ equations. There are two such solutions as far as I can tell. $(x,y,\lambda)=(0,6,0)$ leads to $f(0,6)=0$ so it is a minimum point. $(x,y,\lambda)=(4,2,16\pi)$ leads to $f(0,6)=32\pi$ so it is a maximum point.

Although this method is tedious, it also works for the case when $f$ is a function of more than two variables and there are more than one constraint equation.

 

[chwala]

The goal of the method of Lagrange multipliers is to maximize $f(x,y)$ subject to the constraint $g(x,y)$. Here $f$ and $g$ are assumed to have continuous first partial derivatives. $\lambda$ is a new variable called the Lagrange multiplier. Our goal is to determine the stationary points of the Lagrangian
$$\mathcal{L} (x,y,\lambda)=f(x,y)-\lambda g(x,y)$$
Let $f(x,y)=\pi x^2 y$ and $g(x,y)=x+y-6$. The corresponding Lagrangian is
$$\mathcal{L} (x,y,\lambda)=\pi x^2 y-\lambda (x+y-6)$$
Take partial derivatives as you would compute the gradient of any multi variable function and set it equal to $0$ to find the stationary points.
\begin{align}D_x \mathcal{L}=&2\pi xy-\lambda=0\\
D_y \mathcal{L}=&\pi x^2-\lambda=0\\
g(x,y)=& x+y-6=0\end{align}
Then we look for all the solutions to the system of $3$ equations. There are two such solutions as far as I can tell. $(x,y,\lambda)=(0,6,0)$ leads to $f(0,6)=0$ so it is a minimum point. $(x,y,\lambda)=(4,2,16\pi)$ leads to $f(0,6)=32\pi$ so it is a maximum point.

Although this method is tedious, it also works for the case when $f$ is a function of more than two variables and there are more than one constraint equation.

Cheers Docnet...noted.