Mathematics » Differential Calculus » Applications Of Differential Calculus

# Optimisation Problems

## Optimisation Problems

We have seen that differential calculus can be used to determine the stationary points of functions, in order to sketch their graphs. Calculating stationary points also lends itself to the solving of problems that require some variable to be maximised or minimised. These are referred to as optimisation problems.

The fuel used by a car is defined by $$f(v)=\cfrac{3}{80}{v}^{2}-6v+245$$, where $$v$$ is the travelling speed in $$\text{km/h}$$.

What is the most economical speed of the car? In other words, determine the speed of the car which uses the least amount of fuel.

If we draw the graph of this function we find that the graph has a minimum. The speed at the minimum would then give the most economical speed.

We have seen that the coordinates of the turning point can be calculated by differentiating the function and finding the $$x$$-coordinate (speed in the case of the example) for which the derivative is $$\text{0}$$.

${f}'(v)=\cfrac{3}{40}v-6$

If we set $${f}'(v)=0$$ we can calculate the speed that corresponds to the turning point:

\begin{align*} {f}'(v) & = \cfrac{3}{40}v-6 \\ 0 & = \cfrac{3}{40}v-6 \\ v & = \cfrac{6\times 40}{3} \\ & = 80 \end{align*}

This means that the most economical speed is $$\text{80}\text{ km/h}$$.

### Finding the optimum point:

Let $$f'(x) = 0$$ and solve for $$x$$ to find the optimum point.

To check whether the optimum point at $$x = a$$ is a local minimum or a local maximum, we find $$f”(x)$$:

• If $$f”(a) < 0$$, then the point is a local maximum.

• If $$f”(a) > 0$$, then the point is a local minimum.

## Example

### Question

The sum of two positive numbers is $$\text{10}$$. One of the numbers is multiplied by the square of the other. If each number is greater than $$\text{0}$$, find the numbers that make this product a maximum.

Draw a graph to illustrate the answer.

### Analyse the problem and formulate the equations that are required

Let the two numbers be $$a$$ and $$b$$ and the product be $$P$$.

\begin{align*} a+b &=10 \ldots \ldots (1) \\ P &= a \times b^{2} \ldots \ldots (2) \end{align*}

Make $$b$$ the subject of equation ($$\text{1}$$) and substitute into equation ($$\text{2}$$):

\begin{align*} P &= a(10-a)^{2} \\ &=a ( 100 – 20a + a^{2} ) \\ \therefore P(a) &= 100a – 20a^{2} + a^{3} \end{align*}

### Differentiate with respect to $$a$$

${P}'(a)= 100 – 40a + 3a^{2}$

### Determine the stationary points by letting $$P'(a) = 0$$

We find the value of $$a$$ which makes $$P$$ a maximum:

\begin{align*} {P}'(a) & = 3a^{2} -40a + 100 \\ 0 &= (3a – 10)(a – 10) \\ \therefore a = 10 & \text{ or } a = \cfrac{10}{3} \end{align*}

Substitute into the equation ($$\text{1}$$) to solve for $$b$$:

\begin{align*} \text{If } a = 10 : \enspace b & = 10 – 10 \\ & = 0 \enspace (\text{but } b > 0 ) \\ \therefore \text{no solution} \\ \\ \text{If } a = \cfrac{10}{3} : \enspace b & = 10 – \cfrac{10}{3} \\ & = \cfrac{20}{3} \end{align*}

### Determine the second derivative $$P”(a)$$

We check that the point $$\left(\cfrac{10}{3};\cfrac{20}{3}\right)$$ is a local maximum by showing that $${P}”\left(\cfrac{10}{3}\right) < 0$$:

\begin{align*} {P}”(a) & = 6a -40 \\ \therefore {P}”\left(\cfrac{10}{3}\right) & = 6\left(\cfrac{10}{3}\right) -40 \\ & = 20 -40 \\ & = -20 \end{align*}

The product is maximised when the two numbers are $$\cfrac{10}{3}$$ and $$\cfrac{20}{3}$$.

### Draw a graph

To draw a rough sketch of the graph we need to calculate where the graph intersects with the axes and the maximum and minimum function values of the turning points:

Intercepts:

\begin{align*} P(a) &= a^{3} -20a^{2} + 100a \\ &= a(a – 10)^{2} \\ & \\ \text{Let } P(a) = 0: \enspace & (0;0) \text{ and } (10;0) \end{align*}

Turning points:

\begin{align*} P'(a) & = 0 \\ \therefore a = \cfrac{10}{3} & \text{ or } a = 10 \end{align*}

Maximum and minimum function values:

\begin{align*} \text{Substitute } ( \cfrac{10}{3}; \cfrac{20}{3} ): \enspace P &= ab^{2} \\ &= ( \cfrac{10}{3} ) ( \cfrac{20}{3} )^{2} \\ &= \cfrac{4000}{27} \\ & \approx 148 \qquad (\text{Maximum turning point})\\ & \\ \text{Substitute } (0;10): \enspace P &= ab^{2} \\ &= ( 10 ) ( 0 )^{2} \\ &= 0 \qquad (\text{Minimum turning point}) \end{align*}

Note: the above diagram is not drawn to scale.

## Example

### Question

Michael wants to start a vegetable garden, which he decides to fence off in the shape of a rectangle from the rest of the garden. Michael has only $$\text{160}\text{ m}$$ of fencing, so he decides to use a wall as one border of the vegetable garden. Calculate the width and length of the garden that corresponds to the largest possible area that Michael can fence off.

### Examine the problem and formulate the equations that are required

The important pieces of information given are related to the area and modified perimeter of the garden. We know that the area of the garden is given by the formula:

$\text{Area }= w \times l$

The fencing is only required for $$\text{3}$$ sides and the three sides must add up to $$\text{160}\text{ m}$$.

$160 = w + l + l$

Rearrange the formula to make $$w$$ the subject of the formula:

$w=160 -2 l$

Substitute the expression for $$w$$ into the formula for the area of the garden. Notice that this formula now contains only one unknown variable.

\begin{align*} \text{Area } &= l(160 – 2l) \\ &= 160l – 2l^{2} \end{align*}

### Differentiate with respect to $$l$$

We are interested in maximising the area of the garden, so we differentiate to get the following:

\begin{align*} \cfrac{dA}{dl} = A’ &= 160 – 4l \end{align*}

### Calculate the stationary point

To find the stationary point, we set $${A}'(l)=0$$ and solve for the value(s) of $$l$$ that maximises the area:

\begin{align*} {A}'(l) & = 160 – 4l \\ 0 &= 160 – 4l \\ 4l &= 160 \\ \therefore l &= 40 \end{align*}

Therefore, the length of the garden is $$\text{40}\text{ m}$$.

Substitute to solve for the width:

\begin{align*} w & = 160 -2l \\ & = 160 -2(40 ) \\ & = 160 – 80 \\ & = 80 \end{align*}

Therefore, the width of the garden is $$\text{80}\text{ m}$$.

### Determine the second derivative $${A}”(l)$$

We can check that this gives a maximum area by showing that $${A}”(l) < 0$$:

${A}”(l) = -4$

A width of $$\text{80}\text{ m}$$ and a length of $$\text{40}\text{ m}$$ will give the maximum area for the garden.