06 6maxmin2

# Maximum and Minimum Problems when the Function is Unknown

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Recall that Maximum and Minimum Problems involve finding the largest or smallest possible value for some quantity.

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The following procedure can be used when the formula is not given directly.

1 .. Write down a formula for the quantity whose extreme value is required.
… … … … Define any variables involved.
… … … … A diagram labelled with these variables may also be useful.

2 .. Count the number of variables this formula involves.
… … … … If there is only one variable, then go to step 5
… … … … If there are two or more variables, go to step 3

3 .. Find a relationship between the variables.
… … … … This step is often the most difficult, see below.

4 .. Use this relationship to eliminate all but one variable from the first formula.

5 .. Differentiate and set equal to zero to find the stationary points.

6 .. Determine the nature of the stationary points by using a sign diagram.

7 .. Find the value of the endpoints if the domain is limited

8 .. Determine which point gives the required extreme value.

9 .. Answer the questions in terms of the situation using correct units .

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… … … Step 3 is often the most difficult. Consider the following:

… … … … ☺ Pythagoras’ Theorem

… … … … ☺ Formulas for surface area or volume

… … … … ☺ Similar triangles

… … … … ☺ Trigonometric ratios

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### Example 1

Find the volume of the largest cone that can be inscribed in a sphere of radius 9 cm.

Solution:

… … Let r be the radius of of the cone

… … Let h be the height of the cone.
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… … Volume of a cone $=\dfrac{\pi}{3}r^2h$
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… … From the diagram, we can see that:

… … … $0 \leqslant r \leqslant 9$

… … … $0 \leqslant h \leqslant 18$
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… … Since we have two variables, r and h, we need to find a relationship between them.

… … A labelled diagram of the problem will assist.

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… … We can see a right-angled triangle, so use Pythagoras.

… … … $r^2+(h-9)^2=9^2$

… … … $r^2=9^2-(h-9)^2$

… … … … $=81-(h^2-18h+81)$

… … … … $=18h - h^2$
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… … We can now express the volume of the cone, V, in terms of one variable, h.

… … … $V = \dfrac{\pi}{3} r^2 h$

… … … $V = \dfrac{\pi}{3}(18h-h^2)h \quad \text{ where } 0 \leqslant h \leqslant 18$
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… … To find maximum volume, differentiate V with respect to h and equate to zero.

… … … $\dfrac{dV}{dh}=12{\pi}h-{\pi}h^2$

… … … $12\pi h - \pi h^2 = 0$

… … … $\pi h(12-h)=0$

… … … $h = 0, \;\;or\;\; h = 12$

… … Since a cone with zero height has no volume, reject $h = 0$.

… … We can also ignore the other endpoint, $h = 18$,
… … … because that would give $r = 0$, and a cone with zero radius has no volume.
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… … Gradient table, for $h = 12$,

… … Hence, when $h = 12$, V is maximum.
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… … Calculate maximum volume

… … … $V(12) =\dfrac{\pi}{3}(18\times12-12^2)\times12$

… … … $V(12)=288 \, \pi \; cm^3$
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… … So cone with max volume in a sphere of radius 9cm has:

… … … $h = 21$ cm

… … … $r = 6 \sqrt{2}$ cm

… … … $V =288 \, \pi$ cm3

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### Example 2

For the graph $y=\sqrt{x}$

a) .. Find the coordinates of the point on the graph closest to the point $(4, \; 0)$.

b) .. The distance between the two points.

Solution:

… … Let L be the distance between $(4, \; 0)$ and the point $(x, \; y)$ on the curve.

… … Draw a diagram of the problem

… … The length L can be expressed using Pythagoras.

… … … $L=\sqrt{(4 - x)^2+(y-0)^2}$
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… … We have been given a relationship between x and y.

… … … $y=\sqrt{x} \qquad \text{ for } x \geqslant 0$
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… … We can now express L in terms of the one variable, x.

… … … $L=\sqrt{(4 - x)^2+(\sqrt{x})^2}$

… … … $L =\sqrt{(4 - x)^2+x} \qquad \text{ for } x \geqslant 0$
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… … Now differentiate L with respect to x and equate to zero.

… … … $\dfrac{dL}{dx}=\dfrac{1}{2} \big( (4-x)^2+x \big) ^{-\frac{1}{2}} \big( -2(4-x)+1 \big)$

… … … … . $=\dfrac{2x-7}{2\sqrt{(4-x)^2+x}}$

… … … Let $\dfrac{dL}{dx} = 0$ to find stationary point

… … … $2x-7=0$

… … … $x=\dfrac{7}{2}$
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… … Gradient table at $x=\dfrac{7}{2}$

… … Hence $x=\dfrac{7}{2}$ gives a minimum value for L.
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… … Calculate minimum length of L

… … … $L \Big( \dfrac{7}{2} \Big) = \dfrac{\sqrt{15}}{2}$
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… … The point on the graph of $y = \sqrt{x}$ closest to $(4, \; 0)$

… … is $\left( \dfrac{7}{2}, \; \dfrac{\sqrt{14}}{2} \right)$ … … (a)

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… … The distance between these points is $\dfrac{\sqrt{15}}{2}$ … … (b)

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