06 5maxmin1

# Maximum and Minimum Problems when the Function is Known

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

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

1 .. Differentiate the function

2 .. Set the derivative equal to zero to find the stationary points.

3 .. Determine the nature of the stationary points by using a sign diagram (gradient table).

4 .. Determine which stationary point gives the required extreme value (max or min).

5 .. If domain is restricted, evaluate the value of the quantity at the endpoints.

6 .. Once the absolute maximum or minimum has been established, answer the actual question.

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### Example 1

Let $f:[-1, \; 2] \rightarrow R, f(x) = 4 - x^2$.

a) … Find the absolute maximum and absolute minimum value of the function

b) … Hence state the range.

Solution:

i) … Locate stationary points

… … $f(x) = 4 - x^2$

… … $f'(x) = -2x$

… … stationary point occurs when $f'(x) = 0$

… … $-2x=0$

… … $\;\; x=0$
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… … $f(0) = 4 - 0^2$ … … $f(0) = 4$
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ii) … Determine the nature of stationary points

… … Gradient table for $x = 0$

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

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iii) … Locate endpoints

… … $f(x) = 4 - x^2 \qquad \text{ for } x \in [-1, \; 2]$
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… … $f(-1) = 4 - (-1)^2$ … … $f(-1) = 3$

… … $f(2) = 4 - 2^2$ … … … … $f(2) = 0$
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iv) … Hence answer question (a) Find absolute max and min

… … 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}
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b) … Hence state the range.

… … The range is $y \in \big[ 0,\; 4 \big]$

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### 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 modeled by the formula:

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

… … 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

b) … the minimum number of bacteria

Solution:

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

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

… … Using the CAS calculator:

… … a minimum stationary point has been found at $t = 0.04796$ hours.

… … When $t = 0.04796, \; N = 0.0996782$.

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… … Consider the endpoints

… … … When $t = 0, \; N = 0.1$

… … … When $t = 2, \; N = 3.68$

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… … The absolute minimum occurs at $t = 0.04796$ hours

… … … $60 \times t = 2.8776$ minutes

… … … $60 \times 0.8776 = 52.66$ seconds

… … The least number of bacteria occurs after 2 minutes and 53 seconds.

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b) … the minimum number of bacteria

… … The absolute minimum is at $N = 0.0996782$

… … But N is measured in 100,000 bacteria

… … … $100,000 \times N = 9967.82$

… … … The number of bacteria never drops to 9967, so we round up.

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

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