5 · IB Maths
Optimisation
A rectangle sits under the parabola y = 4 − x². Its area rises, peaks and then falls, and calculus finds the exact width that gives the maximum.
Predict first
Start from the default settings (x = 0.800). What is the sign of the derivative A′(x) if you move the Half-width, x slider to 1.800?
A′(x) = 8 − 6x² is zero at x = 2/√3 ≈ 1.155 and negative beyond it. At x = 1.8, A′ = 8 − 6(3.24) = −11.4, so the area is falling.
Check it in the model: Move the Half-width, x slider from 0.800 to 1.800 and read the A′(x) line in the readouts.
parabola and A(x) rectangle and A′(x) tangent and maximum · drag in any panel to change x
A(x) = 2x(4 − x²) = 8x − 2x³
A′(x) = 8 − 6x²
A′(x) = 0 → x = 2 ÷ √3
A″(x) = −12x
Try this
- Press “Too thin”, “The sweet spot” and “Too wide” in turn and compare the areas. Which setting gives the largest, and what is its percentage of the maximum?
- Tick “Show the derivative A′(x)” and drag x slowly from 0.1 to 1.9. What is the sign of A′(x) before and after the maximum, and what is its value at the top?
- Press “Snap to the maximum” and check that the tangent is horizontal. Then confirm that the exact x is 2/√3 by solving 8 − 6x² = 0.
Exam tip: write the quantity to be optimised as a function of one variable, differentiate, set the derivative to zero, and then show it is a maximum, for example by finding that A″(x) is negative at that value.
More on 5: Chain and product rules · Second derivative and points of inflexion · Trapezoidal rule and Riemann sums · Displacement and distance travelled · Euler's method and differential equations · Implicit differentiation · Related rates of change · Area under a curve · Calculus of motion · Limits · Maclaurin series · Newton–Raphson method (enrichment) · Slope fields · Tangents and the derivative · Volume of revolution