Solving for Optimal Cylinder Dimensions

This page continues the optimization problem for the cylindrical water tank. It shows the graphical representation of the target function and the calculated solution.

Highlight: The optimal dimensions are calculated using calculus and graphical methods.

The graph displays the function y = x - x² + 2000000/x, which represents the surface area in relation to the radius.

Key results:

• Optimal radius $r$ ≈ 68.3 cm
• Optimal height $h$ ≈ 68.2 cm

Example: The minimum point on the graph $68.278, 43937.757$ represents the optimal dimensions for the cylinder.

The solution demonstrates that for minimum surface area, the height and diameter of the cylinder should be approximately equal.

Determining Polynomial Functions

This page introduces Steckbriefaufgaben ganzrationale Funktionen $profile tasks for polynomial functions$. The task is to determine a quadratic function that passes through three given points: $1,3$, $-1,2$, and $3,2$.

Definition: A polynomial function of degree 2 has the general form f$x$ = ax² + bx + c.

The solution process involves:

1. Setting up a system of equations using the given points.
2. Solving for the coefficients a, b, and c.

Example: For the point $1,3$, the equation is a$1$² + b$1$ + c = 3.

The page shows the system of equations: a + b + c = 3 a - b + c = 2 9a + 3b + c = 2

This system will be solved to find the coefficients of the quadratic function.

Solving for Polynomial Coefficients

This page demonstrates the solution to the system of equations for determining the quadratic function.

Highlight: The solution is obtained using a computer algebra system $CAS$.

The CAS solves the system of equations: a + b + c = 3 a - b + c = 2 9a + 3b + c = 2

Result: a = -1 b = 1/2 c = 7/2

Example: The resulting quadratic function is f$x$ = -1/2x² + 1/2x + 7/2.

The page also includes a verification step, checking that the function passes through the given points: f$-1/2$ = 2 f$1$ = 3 f$3$ = 2

This confirms that the derived function satisfies all the given conditions.

Verifying the Polynomial Function

This page shows additional verification steps for the derived quadratic function.

Highlight: The function f$x$ = -1/2x² + 1/2x + 7/2 is confirmed to pass through all given points.

The CAS is used to define the function and evaluate it at the given x-values:

f$-1$ = 2 f$1$ = 3 f$3$ = 2

These results match the original points, confirming the correctness of the solution.

Example: f$3$ = -1/2$3$² + 1/2$3$ + 7/2 = 2

The page also includes an alternative representation of the system of equations using variables x, y, and z, which yields the same solution.

Higher Degree Polynomial Functions

This page introduces the concept of higher degree polynomial functions, specifically cubic functions.

Definition: A cubic function has the general form f$x$ = ax³ + bx² + cx + d.

The page shows the general form of a cubic function and begins to set up an example problem, likely involving finding a cubic function that passes through specific points.

Vocabulary: In Steckbriefaufgaben $profile tasks$, the goal is to reconstruct a function based on given information about its behavior or points it passes through.

The setup suggests that the next steps will involve solving a system of equations to determine the coefficients of a cubic function.

Solving for Cubic Function Coefficients

This page presents a problem to find a cubic function that passes through the points $1,1$, $2,3$, and $3,7$.

Example: For a cubic function f$x$ = ax³ + bx² + cx + d passing through $1,1$, we get the equation a$1$³ + b$1$² + c$1$ + d = 1.

The system of equations is set up: a + b + c + d = 1 8a + 4b + 2c + d = 3 27a + 9b + 3c + d = 7

Highlight: The solution process involves solving this system of equations to find the coefficients a, b, c, and d.

The page shows the initial steps of solving this system, including some algebraic manipulations to simplify the equations.

Complex Polynomial Function Problem

This page presents a more complex problem involving a polynomial function of degree 3.

Highlight: The problem involves finding a cubic function with specific properties, including passing through certain points and having a horizontal tangent at a given point.

Given information:

• The function passes through the point H$0,0$
• It has a horizontal tangent at T$2,-4$
• The function passes through the point B$4,5$

Vocabulary: A horizontal tangent occurs where the derivative of the function equals zero.

The general form of the function is given as: f$x$ = ax³ + bx² + cx + d

The page sets up the system of equations based on these conditions:

1. f$0$ = 0
2. f$2$ = -4
3. f'$2$ = 0
4. f$4$ = 5

This system will be solved to determine the coefficients of the cubic function.

Solving Complex Polynomial System

This page continues the solution process for the complex cubic function problem.

The system of equations is fully set up:

1. d = 0 $from f(0$ = 0)
2. 8a + 4b + 2c = -4 $from f(2$ = -4)
3. 12a + 4b + c = 0 $from f'(2$ = 0)
4. 64a + 16b + 4c = 5 $from f(4$ = 5)

Example: The equation 12a + 4b + c = 0 comes from setting the derivative f'$x$ = 3ax² + 2bx + c equal to zero at x = 2.

The page shows the solution process using a computer algebra system:

$8a + 4b + 2c = -4 12a + 4b + c = 0$ solved for a and b

Result: a = 1, b = -3

Highlight: The final cubic function is determined to be f$x$ = x³ - 3x² + 0.25x.

This solution satisfies all the given conditions, including passing through the specified points and having a horizontal tangent at T$2,-4$.

Page 10: Homework Assignment

Concludes with homework assignments and reference to additional practice problems.

Highlight: References specific textbook exercises for further practice.

Optimizing Cylinder Dimensions

This page presents a problem involving the optimization of a cylindrical water tank's dimensions. The goal is to minimize the surface area while maintaining a volume of 1000 liters.

Highlight: The problem focuses on finding the optimal dimensions of a cylindrical water tank without a lid, with a volume of 1000 liters, to minimize material usage.

The solution process begins with setting up the volume equation:

πr²h = 1,000,000 cm³

The surface area function is then derived:

O$r$ = πr² + 2πrh

By substituting h from the volume equation and simplifying, we get the target function:

O$r$ = πr² + 2,000,000/r

Example: The target function O$r$ = πr² + 2,000,000/r represents the surface area in terms of the radius.

The page shows the initial steps of the optimization process, including setting up the equations and preparing for differentiation.