# Find the greatest and least values of Volume of cylinder

• chwala

#### chwala

Gold Member
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...

Last edited:
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.

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?

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.

• 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.

• 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.

• docnet