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:
$hequals;\left\{0h5\right\}comma;h\in \mathrm{reals;}V\left(h\right)equals;$
$h\cdot \left(20-2h\right)\cdot \left(10-2h\right)$${}$
$4\mathbf{}{h}^{3}-60{h}^{2}plus;200h$
First derivative must be found to find a x value that minimizes T
$\frac{\ⅆV}{\ⅆ\mathrm{h}}$
$\=$$12{h}^{2}-120hplus;200$
Set the derivative to 0
$0\=$
$12{h}^{2}-120hplus;200$
$h\=$
$5\+\frac{5}{3}\mathbf{}\sqrt{3}comma;5-\frac{5}{3}\mathbf{}\sqrt{3}$
$hequals;$
$2.113248653\,7.886751347$
As h = 7.886751347 is outside the limit, only h = 2.113248653 and the end points should be tested
$V\left(0\right)equals;0{\mathrm{cm}}^{3}$
$V\left(2.113248653\right)equals;192.4500897{\mathrm{cm}}^{3}$
$V\left(5\right)equals;0{\mathrm{cm}}^{3}$
Hence when h ≈ 2.11cm a maximum volume can be achieved.
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