Optimization: A Volume Example - Maple Help

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Optimization: A Volume Example

Main Concept

An optimization problem involves finding the best solution from all feasible solutions. One is usually solving for the largest or smallest value of a function, such as the shortest distance or the largest volume.

A minimum or maximum of a continuous function over a range must occur either at one of the endpoints of the range, or at a point where the derivative of the function is 0 (and thus the tangent line is horizontal). These are called critical points.

Steps

1. 

Identify what value is to be maximized or minimized.

2. 

Define the constraints.

3. 

Draw a sketch or a diagram of the problem.

4. 

Identify the quantity that can be adjusted, called the variable, and give it a name, such as h.

5. 

Write down a function expressing the value to be optimized in terms of h.

6. 

Differentiate the equation with respect to h.

7. 

Set the equation to 0 and solve for h.

8. 

Check the value of the function at the end points.

 

Problem: Alice is given a piece of cardboard that is 20cm by 10cm. She wants to make an open top box by cutting the corners and folding up the sides.

 

Let h be the height of the box. Adjust the value of h using the slider to find the value that maximizes the volume.

 

 

 

 

 

h =

Numerical solution

 

Volume of the box is given by:

h &equals;  0<h< 5&comma; h&reals;       Vh  &equals;

 h202 h102 h

4h360 h2&plus; 200 h

First derivative must be found to find a x value that minimizes T

&DifferentialD;V&DifferentialD; h 

&equals; 12 h2  120 h &plus;200

Set the derivative to 0

0&equals;

12 h2  120 h &plus;200

 

h&equals; 

5&plus;533 &comma; 5533

h &equals; 

2.113248653&comma; 7.886751347

As h = 7.886751347 is outside the limit, only h = 2.113248653 and the end points should be tested

V0 &equals; 0 cm3

V2.113248653 &equals; 192.4500897 cm3

V5 &equals; 0 cm3

Hence when h ≈ 2.11cm a maximum volume can be achieved.

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