Problem Solving with Differentiation (Edexcel IGCSE Maths B): Revision Note

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 $\frac{\mathrm{d}A}{\mathrm{d}t}=0$

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 $\left(10-x\right)$ m

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

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

    • $V=2x\left(10-x\right)$

  • 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 $\frac{\mathrm{d}V}{\mathrm{d}x}=0$

  • Find $\frac{\mathrm{d}V}{\mathrm{d}x}$ by differentiating each term

    • $\frac{\mathrm{d}V}{\mathrm{d}x}=20-4x$

  • Set this equal to zero and solve

    • $\begin{array}{rcl}20-4x& =& 0\end{array}$ 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 $V$

    • $V=20\times 5-2\times 5^2=50$

    • The maximum volume is 50 m³

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 $V=2x\left(10-x\right)$