09Gmaxminwithunknownrule

= Solving Maximum and Minimum Problems = = When the Rule is not known =

If the rule is not given in the question, then follow these steps
 * 1) Draw a diagram of the situation, put in all known information
 * 2) Write an equation linking the given information
 * 3) Identify the quantity to be maximised or minimised
 * 4) Write this quantity as an equation in terms of only one variable
 * 5) Sketching a graph of this equation may help
 * 6) Differentiate
 * 7) Set the derivative to equal zero and solve
 * 8) Draw a gradient table (sign diagram) to determine if you have found a maximum or minimum
 * 9) Check the endpoints of the domain in case they are the required max or min
 * 10) Answer the question being asked


 * Example 1 **

... ... A rectangular frame is to be made from a piece of wire, 200cm long ... ... ** (a) ** .. If the width is x, find an expression for the length ... ... ** (b) ** .. Use this to write an expression for the Area in terms of x (only) ... ... ** (c)  ** .. Find the value of x that gives the maximum area ... ... ** (d)  ** .. Hence find the maximum area and state the dimensions of the rectangle to give the maximum area


 * Solution: **

... ... ** (a) ** .. If the width is x, find an expression for the length


 * Draw a diagram with the known information



math . \qquad \qquad \text{Perimeter } = 2 \times \text{Length} + 2 \times \text{Width} \\.\\ . \qquad \qquad 2L + 2x = 200 \qquad \big\{ \div 2 \big\} \\.\\ . \qquad \qquad \; L + x = 100 \qquad \big\{ - x \big\} \\.\\ . \qquad \qquad \; L = 100 - x math


 * Add this information to your diagram



... ... ** (b)  ** .. Use this to write an expression for the Area in terms of x (only)

math . \qquad \qquad \text{Area } = \text{ Width } \times \text{ Length } \\.\\ . \qquad \qquad A \big( x \big) = x \big( 100 - x \big) \\.\\ . \qquad \qquad A \big( x \big) = 100x - x^2 math

... ... ** (c) ** .. Find the value of x that gives the maximum area


 * Differentiate A(x)

math . \qquad \qquad A '\big( x \big) = 100 - 2x math


 * Set A'(x) = 0 to find stationary point

math . \qquad \qquad 100 - 2x = 0 \\.\\ . \qquad \qquad \;\; -2x = -100 \\.\\ . \qquad \qquad \qquad x = 50 math
 * Draw gradient table to confirm this is a maximum


 * The gradient table shows that a maximum occurs at x = 50 cm


 * Examination of the graph shows the endpoints are not maximums

... ... ** (d) ** .. Hence find the maximum area and state the dimensions

math . \qquad \qquad A \big( x \big) = 100x - x^2 \\.\\ . \qquad \qquad A \big( 50 \big) = 100 \big( 50 \big) - \big( 50 \big)^2 \\.\\ . \qquad \qquad A \big( 50 \big) = 2500 math


 * Hence Maximum Area is 2500 cm 2
 * Dimensions are:
 * Width = 50 cm
 * Length = 50 cm

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