Minimum surface area with fixed volume Find the dimensions of the closed right circular cylindri

Minimum surface area with fixed volume Find the dimensions...

Key Concepts

Calculus Optimization

Calculus optimization involves using techniques such as differentiation to identify the extrema of a function. This usually includes finding the first derivative, setting it equal to zero to locate critical points, and confirming whether these points correspond to minima or maxima. It is fundamental in problems where one seeks to optimize (minimize or maximize) an objective function subject to certain conditions.

Constraint Optimization

Constraint optimization is the process of optimizing an objective function while adhering to one or more constraints. In many problems, a constraint (like a fixed volume) is used to express one variable in terms of another, reducing the problem to a one-variable optimization task. Techniques such as substitution or Lagrange multipliers are commonly employed to handle such constraints effectively.

Geometric Formulation of Cylindrical Properties

This concept involves understanding and applying the standard formulas for the volume and surface area of a right circular cylinder. The volume is typically expressed as a function of the radius and height, while the surface area includes the areas of the two circular ends and the lateral surface. Accurately formulating these expressions is essential when solving optimization problems with geometric figures.

Transcript

00:02 Okay, so we have a cylinder in this question with a volume fixed to be 16 cm cubed.

00:15 I denote the height of the cylinder by age and the radius of the belt by a small r.

00:25 And we have to minimize the surface area of the cylinder with this given volume.

00:43 Volume.

00:45 So actually let's write down what's the volume in terms of r and h.

00:58 This is well known n equals pi r squared times h.

01:14 And the surface area is the 2 times area of the base, so 2 times pi r squared and height times the circumference of the base so 2x r okay so minimize a and the condition that v is equal to 60 so this is our constraint that v is equal to 60 okay so we will do it using lagrange so we need to find r and h, set that gradient of a is equal to lambda times gradient of v and lambda so let's compute the gradient, gradient of a with respect to r is equal 4 pi r plus 2 pi and derivative with respect to h will be just 2 pi r.

03:10 Okay, and derivative of gradient of function v with respect to r equals 2x pi r and pi r and pi r square with respect to h right so we just equal these two coordinates.

03:46 So we have system of equations for pi r just 2 pi h equals lambda times 2 pi r and 2 pi r equals lambda pi r squared and we have to remember about our a constraint which is pi r squared h equals actually i should write here the other constraint is pi r squared times h minus 16 equals zero so if you prefer call it vitar style or something just to make sure it's not volume as given a book all right so so we have our system here and we have some.

05:12 Let's do some simplifications.

05:14 We can divide here both sides by pi and by 2 actually.

05:23 You will have 2...

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