5.7 OptimisationIB Maths: Applications and Interpretation SL: Revision notes

Optimisation means finding the maximum or minimum value of a quantity in a real situation. The method is the same each time:

1. Choose a variable and write the quantity to be optimised as a function of one variable, using any given constraint to eliminate others.
2. State the domain from the context (lengths and amounts cannot be negative).
3. Find where the derivative is zero, by hand or by using your GDC to graph the function or its gradient.
4. Check whether the stationary point is a maximum or minimum (sign of the gradient either side, or the GDC graph) and compare with the ends of the domain.
5. Answer in context, with units.

Profit is revenue minus cost, $P=R-C$. If a profit model is $P(x)=-2x^2+80x-300$ with $x$ items sold, then $P'(x)=-4x+80=0$ gives $x=20$. The gradient changes from positive to negative, so this is a maximum, and $P(20)=500$. For cost, such as $C(x)=2x^2-48x+500$, $C'(x)=4x-48=0$ gives $x=12$ and $C(12)=212$, a minimum because the gradient changes from negative to positive. Always give money to 2 decimal places or as a sensible whole amount, with the currency (USD, AED).

A rectangle has a perimeter of 60 m and width $x$. The length is $30-x$, so $A=x(30-x)=30x-x^2$. $A'(x)=30-2x=0$ gives $x=15$ and $A=225$ m². The best rectangle is a square. The domain is $0<x<30$, because both sides must be positive.

A closed cylinder with surface area 600 cm² has $2\pi r^2+2\pi rh=600$, so $h=\frac{300-\pi r^2}{\pi r}$. Then $V=\pi r^2h=300r-\pi r^3$ and $\frac{dV}{dr}=300-3\pi r^2=0$ gives $r=5.64$ cm. Test the gradient either side: $V'(5)=64.4>0$ and $V'(6)=-39.3<0$, so it is a maximum, with $V=1130$ cm³ and $h=11.3$ cm. For an open box made from a 30 cm square card by cutting squares of side $x$ from the corners, $V=x(30-2x)^2$, which is greatest at $x=5$, giving 2000 cm³.

Graph the model on your GDC with a window that matches the domain. Use the maximum or minimum function to read the turning point, or graph $f'(x)$ and find where it crosses the $x$-axis. Either approach gives the stationary value of $x$; then evaluate the function to find the optimum value. Examination questions often say “use your GDC”, so write down the equation or the graph you used, and give answers exact or to 3 significant figures.

A stationary point is not automatically the answer. Reject solutions outside the domain: for $V'(x)=12(x-5)(x-15)=0$ on $0<x<15$, $x=15$ is rejected. Compare with the ends of the domain: if a factory must produce between 15 and 30 tonnes and $C$ is increasing from $x=12$, the least cost is at $x=15$, not at the stationary point. Finally interpret: say what the value means, such as “the garden has a maximum area of 225 m² when it is a 15 m square”.