08Dmaxminproblemsunknownfunction

Maximum and minimum problems when the function is unknown

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. This may require certain variables to be defined. A diagram labelled with these variables may also be useful.
 * 2) Count the number of variables this formula involves.
 * 3) If there is only one variable, then go to step 4
 * 4) If there are two or more variables, go to step 3
 * 5) Find a relationship between the variables.
 * 6) This step is often the most difficult, see below.
 * 7) Use this relationship to eliminate all but one variable from the first formula.
 * 8) Differentiate and set equal to zero to find the stationary points.
 * 9) Determine the nature of the stationary points by using a sign diagram.
 * 10) Determine which stationary point gives the required extreme value.
 * 11) Answer the questions in terms of the situation using correct units.

Step 3 is often the most difficult. Consider the following: ☺ Pythagoras’ Theorem ☺ Formulas for surface area or volume ☺ Similar triangles ☺ Trigonometric ratios


 * Example 1 **

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


 * __Solution:__**

math . \qquad \text{Volume of a cone}=\dfrac{\pi}{3}r^2h \qquad. math

Let r be the radius of of the cone and let h be the height of the cone.

From the situation, we can see that: math . \qquad 0 < r \leqslant 9 \qquad. \\ . \\ . \qquad 0 < h \leqslant 18 \qquad. math

Since we have two variables, r and h, we need to find a relationship between them.

A labelled diagram of the problem will assist.

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

math \\ . \qquad r^2+(h-9)^2=9^2 \qquad \\. \\ . \qquad r^2=9^2-(h-9)^2 \qquad \\. \\ .\qquad \quad=81-(h^2-18h+81) \qquad. \\ . \\ . \qquad \quad=18h - h^2 math

We can now express the volume of the cone, V, in terms of one variable, h.

math . \qquad V = \dfrac{\pi}{3} r^2 h \\. \\ . \qquad V = \dfrac{\pi}{3}(18h-h^2)h \quad \text{ where } 0 < h \leqslant 18 \qquad. math

To find maximum volume, differentiate V with respect to h and equate to zero.

math . \qquad \dfrac{dV}{dh}=12{\pi}h-{\pi}h^2 \qquad. \\ . \\ . \qquad \quad = \pi h (12 - h) math

math . \qquad {\pi}h(12-h)=0 \qquad \text{ when } \; h = 0, \;12 \qquad. math

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

Show that when h = 12, V is maximum.

When h = 12cm, math \\ . \qquad V=\dfrac{\pi}{3}(18\times12-12^2)\times12 \qquad. \\ . \\ . \qquad V=288 \, \pi \; cm^3 math

So cone with max volume in a sphere of radius 9cm has: math . \qquad h = 21 \text{ cm } \qquad. \\ . \\ . \qquad r = 6 \sqrt{2} \text{ cm } \qquad .\\. \\ . \qquad V =288 \, \pi \; cm^3 \qquad. math


 * Example 2 **

math \text{For the graph } y=\sqrt{x} \qquad. math ... ... 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 graph of the problem which displays the distance, L, which needs to be minimised and the point (x, y) on the curve.

For L to be minimum, the distance between (x, y) and (4, 0) must be minimum.

The length L can be expressed using Pythagoras.

math . \qquad L=\sqrt{(x-4)^2+(y-0)^2} \qquad. math

We have been given a relationship between x and y.

math . \qquad y=\sqrt{x} \qquad. math

We can now express L in terms of x. math \\ . \qquad L=\sqrt{(x-4)^2+(\sqrt{x})^2} \qquad. \\ . \\ . \qquad \quad=\sqrt{(x-4)^2+x} math

Now differentiate L with respect to x and equate to zero. math \\ . \qquad \dfrac{dL}{dx}=\dfrac{1}{2} \big( (x-4)^2+x \big) ^{-\frac{1}{2}} \big( 2(x-4)+1 \big) \qquad .\\ .\\ . \qquad \quad=\dfrac{2x-7}{2\sqrt{(x-4)^2+x}} math

math \\ . \qquad \dfrac{dL}{dx} = 0 \qquad \text{ when } \qquad 2x-7=0 \qquad. \\ . \\ . \qquad \Longrightarrow \quad x=\dfrac{7}{2} math

math \text{The point closest to } (4, \; 0) \text{ is } \left( \dfrac{7}{2}, \; \dfrac{\sqrt{14}}{2} \right) \qquad. math

math \text{The distance between these points is } L \left( \dfrac{7}{2} \right) = \dfrac{\sqrt{15}}{2} \qquad. math

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