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
Identify what value is to be maximized or minimized.
Define the constraints.
Draw a sketch or a diagram of the problem.
Identify the quantity that can be adjusted, called the variable, and give it a name, such as h.
Write down a function expressing the value to be optimized in terms of h.
Differentiate the equation with respect to h.
Set the equation to 0 and solve for h.
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 = 0<h< 5, h∈ℝ Vh =
h⋅20−2 h⋅10−2 h
4h3−60 h2+ 200 h
First derivative must be found to find a x value that minimizes T
ⅆVⅆ h
= 12 h2 − 120 h +200
Set the derivative to 0
0=
12 h2 − 120 h +200
h=
5+533 , 5−533
h =
2.113248653, 7.886751347
As h = 7.886751347 is outside the limit, only h = 2.113248653 and the end points should be tested
V0 = 0 cm3
V2.113248653 = 192.4500897 cm3
V5 = 0 cm3
Hence when h ≈ 2.11cm a maximum volume can be achieved.
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