Solution

Since all turning points are stationary points, we can use the method for finding stationary points to find the turning points of this function:

1) The first derivative of the function $y=x^3-10x^2+25x+4$ with respect to $x$ is: \[\dfrac{\mathrm{d} y}{\mathrm{d} x}=3x^2-20x+25\]

2) Setting this equal to zero gives: \[3x^2-20x+25=0\]

3) We must now solve the above equation for $x$. Using the quadratic formula with $a=3$, $b=-20$ and $c=25$ gives $x=5$ or $x=\frac{5}{3}$. There is a stationary point at $x=5$ and $x=\frac{5}{3}$. The corresponding $y$ values at each of these points are found by substituting each value of $x$ into the function:

• For $x=5$, we have

\begin{align} y&=5^3-10\times 5^2+25\times 5+4\\ &=125-250+125+4\\ &=4 \end{align}

• For $x=\dfrac{5}{3}$, we have

\begin{align} y&=\left(\frac{5}{3}\right)^3-10\times \left(\frac{5}{3}\right)^2+25\times \frac{5}{3}+4\\ &=\frac{608}{27} \end{align} So the stationary points of the function $y=x^3-10x^2+25x+4$ are at and $(5,4)$ and $\left(\dfrac{5}{3},\dfrac{608}{27}\right)$.

4) For this example we will use the second derivative test to determine whether each of these stationary points is a maximum, a minimum, or a point of inflection.

We find the second derivative of the function $y=x^3-10x^2+25x+4$ by differentiating the first derivative, $\dfrac{\mathrm{d} y}{\mathrm{d} x}=3x^2-20x+25$, with respect to $x$: \[\dfrac{\mathrm{d}^2 y}{\mathrm{d} x^2}=6x-20\] We must now look at the sign of the second derivative at each of the stationary points that we have found. When:

• $x=\frac{5}{3}$ the second derivative is:

\begin{align} \dfrac{\mathrm{d}^2 y}{\mathrm{d} x^2}&=6x-20\\ &=6\times \frac{5}{3}-20\\ &=-10 \end{align} Since $(-10)$ is less than zero, the second derivative is negative at this stationary point and so we have found a maximum.

• $x=5$ the second derivative is:

\begin{align} \dfrac{\mathrm{d}^2 y}{\mathrm{d} x^2}&=6x-20\\ &=6\times 5-20\\ &=10 \end{align} Since $10$ is greater than zero, the second derivative is positive at this stationary point and so we have found a minimum.

Now, the original function, $y=x^3-10x^2+25x+4$, is a cubic function. All cubic functions have the shape of an “S” tilted on its side. When:

• The coefficient in front of the $x^3$ term is positive, as $x$ tends towards positive infinity (becomes larger and larger), $y$ also tends towards positive infinity, while as $x$ tends towards minus infinity (becomes more and more negative), $y$ also tends towards minus infinity, so the graph will look like this:

• The coefficient in front of the $x^3$ term is negative, as $x$ tends towards positive infinity (becomes larger and larger), $y$ tends towards negative infinity, while as $x$ tends towards minus infinity, $y$ tends towards positive infinity, so the graph will look like this:

In either case, we can see that any turning points of a cubic function can only be local maxima or minima. The global maximum and minimum of a cubic function do not exist.

The maximum and minimum points of the function $y=x^3-10x^2+25x+4$ found above are therefore the local maximum and minimum points of the function.

Solution

To determine the quantity $q$ of calculators which maximises profit we must first find the derivative of the profit function $\pi (q)=100q-0.1q^2$:

\[\pi '(q)=100-0.2q\]

We must then set this derivative equal to zero and solve for $q$ as follows: \begin{align} 100-0.2&q=0\\ \Rightarrow 0.2&q=100\\ \Rightarrow &q=500 \end{align}

So the function $\pi (q)=100q-0.1q^2$ has a turning point when $q=500$, but is this a maximum? To answer this question we will use the second derivative test. To find the second derivative of the function we must differentiate the first derivative. This gives: \[\pi ''(q)=-0.2\]

As the second derivative does not depend on $q$ (it is a constant), it will be equal to $-0.2$ (negative) for all values of $q$. As $-0.2$ is negative, we have found a maximum and so the company should sell $500$ calculators per day to maximise profit. Another way to see that $q=500$ is a maximum is to plot the function $\pi (q)=100q-0.1q^2$:

We can see from the graph that the function $\pi (q)=100q-0.1q^2$ takes the highest value when $q=500$ and that the slope changes from being positive when $q\lt500$ to negative when $q\gt500$.

We now wish to find the maximum daily profit for the company. To do this we substitute the profit-maximising quantity of calculators sold, $q=500$, into the profit equation $\pi (q)=100q-0.1q^2$:

\begin{align} \pi (5)&=100×500-0.1×(500)^2\\ &=50,000-25,000\\ &=25,000\\ \end{align}

so the company’s maximum daily profit is $£25,000$.

Solution

a) A firm's total costs are made up of its fixed and its variable costs. Fixed costs are constant (do not depend on the quantity of output produced) while variable costs vary with output. The constant term in the bakery's total cost function is $125$ so the bakery's fixed costs are $125$. The bakery's variable costs must therefore be given by: \[5q^2-40q+125-125=5q^2-40q\].

To find the value of $q$ which minimises the firm's variable costs, we must first differentiate $5q^2-40q$ with respect to $q$. This gives $10q-40$. We then need to set this derivative equal to zero and solve for $q$: \begin{align} 10q-40&=0\\ \Rightarrow 10q&=40\\ \Rightarrow q&=4 \end{align}

So the firm's variable costs are minimised when the firm produces $4$ loaves of bread.

b) We can obtain a firm's marginal cost (MC) function by differentiating the total cost function with respect to $q$. For the given total cost function, this gives: \begin{align} MC(q)&=TC'(q)\\ &=10q-40 \end{align}

Notice that this is the same as the derivative of the variable cost function. This makes sense as the derivative of a constant (fixed costs) is zero.

A firm's average cost (AC) is defined as total cost divided by output. Dividing the bakery's total cost function by $q$ gives: \begin{align} AC(q)&=\dfrac{TC(q)}{q}\\ &=\dfrac{5q^2-40q+125}{q}\\ &=5q-40+\dfrac{125}{q} \end{align}

c) To find the value of $q$ which minimises the bakery's average cost, we must first differentiate the bakery's average cost function. This gives: \[AC'(q)=5-\dfrac{125}{q^2}\] We must then set this derivative equal to zero and solve for $q$: \begin{align} 5-&\dfrac{400}{q^2}=0\\ &\Rightarrow 5=\dfrac{125}{q^2}\\ &\Rightarrow q^2=\dfrac{125}{5}\\ &\Rightarrow q^2=25\\ &\Rightarrow q=5\text{ (since }q\text{ cannot be negative)}\\ \end{align} So the bakery's average costs are minimised when $q=5$. Now, when $q=5$ the firm's marginal cost is: \begin{align} MC(5)&=10\times 5-40\\ &=10 \end{align} and the firm's average cost is: \begin{align} AC(5)&=5\times 5-40+\dfrac{125}{5}\\ &=25-40+25\\ &=10 \end{align} So when $q=5$ both the bakery's average cost and its marginal cost are both equal to $10$.

Solution

a) Total revenue is equal to the price of the good multiplied by the quantity sold. Since the subject of the inverse demand function is the market price $p$, we can obtain the total revenue function from the inverse demand function by multiplying the inverse demand function by $q$. This gives: \begin{align} TR(q)&=q\left (-\dfrac{1}{4}q+300\right )\\ &=300q-\dfrac{1}{4}q^2 \end{align}

We can then obtain the marginal revenue function by differentiating the total revenue function with respect to $q$: \begin{align} MR(q)&=TR'(q)\\ &=300-2\times \dfrac{1}{4}q\\ &=300-\dfrac{1}{2}q \end{align}

b) To determine the quantity of lawnmowers which maximises total revenue, we must differentiate the total revenue function, set the derivative equal to zero and solve for $q$: \begin{align} TR'(q)&=300-\dfrac{1}{2}q\;\;\;\text{ (from part}\textbf{ a)}\text{)}\\ \Rightarrow 300&-\dfrac{1}{2}q=0\\ \Rightarrow q&=2\times 300\\ \Rightarrow q&=600 \end{align} So the revenue-maximising quantity of lawnmowers is $600$. From the inverse demand function, when $q=600$ we know that \begin{align} p&=-\dfrac{1}{4}\times 600+300\\ &=-150+300\\ &=150 \end{align} So the revenue-maximising price per lawnmower is $150$.