Problem Solving with Differentiation (Edexcel IGCSE Maths)

Spec point — 3.4E

Problem-solving with differentiation

What is problem-solving using differentiation?

• You can use the same method of differentiating curves to find turning points to help with problems involving finding the maximum or minimum value of a quantity

  • These are called optimisation problems

• Questions may use different variables

  • For example, $A=t^{3}-12t$ and $\frac{dA}{dt}=0$

How do I apply differentiation to different contexts?

• This is easiest explained using an example

• Find the maximum volume, $V$ m³, of the cuboid shown with width 2 m, length $x$ m and height $( 10 - x )$ m

  • Find a formula for its volume in terms of $x$

  • Volume = width $\times$ length $\times$ height

    • $V = 2 x ( 10 - x )$

  • Expand this into individual terms in $x$

    • $V=20x-2x^{2}$
    • This is a negative quadratic in $x$ (it has an $\cap$ shape)

• The graph of $V$ against $x$ will have a maximum point when $\frac{dV}{dx}=0$

  • Find $\frac{dV}{dx}$ by differentiating each term

    • $\frac{dV}{dx}=20-4x$

  • Set this equal to zero and solve

    • $\begin{gathered} 20 - 4 x & = 0 \end{gathered}$ so $4x=20$ giving $x=5$
    • This is the value of $x$ at the maximum point (not the value of $V$)

  • Substitute this value of $x$ back into the equation for $V$

    • $V=20\times5-2\times5^{2}=50$
    • The maximum volume is 50 m³

Exam Hint A common problem in the exam is to forget to substitute the value of $x$ back into the formula!

How do I know if I have found a maximum value or a minimum value?

• Sometimes there are two values of $x$ for different turning points and you need one of them

  • either substitute them both back into the formula

    • See which gives the max value and which gives the min value

  • or use any acceptable techniques for classifying turning points

    • For example, using a sketch to see if its a max or min

  • Remember that

    • Positive quadratics have minimum points
    • Negative quadratics have maximum points

How do I use differentiation if there are lots of variables?

• Sometimes a formula has lots of letters

  • You need to find an extra relationship between these letters
  • then substitute it into the formula

• If, in the above example, you had width 2 m, length $x$ m and height $y$ m then

  • $V=2xy$

    • But you need a formula in $x$ only

  • You will be given an extra piece of information, such as the length and height sum to 10 m

    • Therefore $x+y=10$
    • Make $y$ the subject, $y=10-x$
    • Substitute it into $V$ to get $V = 2 x ( 10 - x )$

Worked Example A farmer has 60 metres of fencing and wants to fence off the biggest rectangular area possible next to an existing wall.

The area has dimensions $x$ metres by $y$ metres, as shown.

(a) Explain why $2x+y=60$.

Answer:

The wall is not part of the fencing

Find the total length of the three sides of the fence shown

$x+y+x$

This must equal 60 metres

The length of the fence must equal 60 metres so$2x+y=60$

(b) Show that the area, $A$m², is given by $A=60x-2x^{2}$.

Answer:

Find the area of the rectangle shown

$A=xy$

You need a right-hand side in terms of $x$ only

Make $y$ the subject of part (a)

$y=60-2x$

Substitute this into the area formula

$A = x ( 60 - 2 x )$

This is now all in terms of $x$

Expand

$A=60x-2x^{2}$

(c) Find the maximum possible area.

Answer:

To find a maximum point, set $\frac{dA}{dx}=0$

First find $\frac{dA}{dx}$ by differentiating each term

$\frac{dA}{dx}=60-4x$

Set this equal to zero and solve

$\begin{gathered} 60 - 4 x & = 0 \\ 4 x & = 60 \\ x = \frac{60}{4} \\ x = 15 \end{gathered}$

This is the value of $x$ at the maximum point (but not the maximum of $A$)

Substitute this value of $x$ back into $A$

$\begin{gathered} A = 60 \times 15 - 2 \times 15^{2} \\ = 450 \end{gathered}$

The maximum area is 450 m²

(d) Explain how you know that the answer in part (c) is a maximum area, not a minimum area.

Answer:

See how the area $A$ depends on $x$

Think about what it would look like as a graph

$A=60x-2x^{2}$ is a negative quadratic curve

A negative quadratic curve has an $\cap$ shape

There is only one turning point on this graph and it is a maximum point

Final answer: The curve$A=60x-2x^{2}$is a negative quadratic curve

Final answer: This means it can only have a maximum point, not a minimum point