08Cmaxminproblemsknownfunction

Maximum and minimum problems when the function is known


 * Many practical problems require some quantity to be minimised, for example the cost of manufacturing.
 * Other quantities need to be maximised, for example, sales profit, area of an enclosure.
 * We can use differential calculus to help solve many of these problems.

When we are provided with an equation of the quantity to be minimised or maximised we can follow the steps below:


 * 1) Differentiate and set equal to zero to find the stationary points.
 * 2) Determine the nature of the stationary points by using a sign diagram (gradient table).
 * 3) Determine which stationary point gives the required extreme value.
 * 4) If domain is restricted, evaluate the value of the quantity at the endpoints.
 * 5) Once the absolute maximum or minimum has been established, answer the actual question.


 * Example 1 **

... ... Let f : [–1, 2] → R, f(x) = 4 – x 2.

... ... Find the absolute maximum and absolute minimum value of the function ... ... and hence state the range.


 * Solution:**

math . \qquad f'(x) = -2x math


 * stationary point occurs when f '(x) = 0

math \\ . \qquad f'(x)=0 \text{ when } \\. \\ . \qquad -2x=0 \\. \\ . \qquad \;\; x=0 math

... ...
 * The gradient table shows that x = 0 is a local maximum.


 * Observation of the graph shows that:
 * The absolute maximum value occurs at x = 0 and is y = 4.
 * The absolute minimum value occurs at x = 2 and is y = 0. {at an endpoint}

math . \qquad \text{Hence, the range is } y \in \big[ 0,\; 4 \big] math


 * Example 2 **


 * The number of a type of bacteria, N(t) .. (N is measured in 100,000 bacteria)
 * in a two hour laboratory experiment is modelled by the formula

math . \qquad N(t)=0.4t^3+0.1t^2+\dfrac{0.8}{t+8} \qquad \text{where } 0 \leqslant t \leqslant 2 math


 * where t is the time in hours from the start of the experiment

... Find ... ... (a) .. the time (in minutes and seconds) when the number of bacteria is at its lowest and ... ... (b) .. the minimum number of bacteria


 * Solution:**

... ... ** (a) ** .. the time (in minutes and seconds) when the number of bacteria is at its lowest

math . \qquad N(t)=0.4t^3+0.1t^2+\dfrac{0.8}{t+8} \quad \text{ where }\quad 0\leqslant t \leqslant 2 math


 * Using the CAS calculator, a minimum stationary point has been found at t = 0.04796 hours.


 * When t = 0.04796, N = 0.09968.


 * Consider the endpoints
 * When t = 0, N = 0.1
 * When t = 2, N = 3.68


 * The absolute minimum occurs at t = 0.04796 hours
 * The least number of bacteria occurs after 2 minutes and 53 seconds.

... ... ** (b) ** .. the minimum number of bacteria


 * The absolute minimum is at N = 0.09968


 * But N is measured in 100,000 bacteria


 * So the least number of bacteria in the first two hours is 9968.

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