Optimisation means finding the largest or smallest value of something, such as area or cost. Write the quantity as a function of one variable, differentiate, set the derivative to zero and solve.
A pen against a wall uses 40 m of fencing on three sides. With width x, the length is 40 − 2x and the area is A = x(40 − 2x) = 40x − 2x². dA/dx = 40 − 4x = 0 gives x = 10, and d²A/dx² = −4 < 0 confirms a maximum. The largest area is 10 × 20 = 200 m².
Worked example
Find the maximum point of y = 2x³ − 18x² + 48x − 2.
- dy/dx = 6x² − 36x + 48 = 6(x − 2)(x − 4) = 0.
- d²y/dx² = 12x − 36: at x = 2 it is −12, a maximum; y = 16 − 72 + 96 − 2 = 38.
Answer:(2, 38)
Key words
- optimise
- find the best (largest or smallest) value
- constraint
- a condition that links the variables
- objective function
- the quantity being maximised or minimised
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