You cut equal squares from the corners of a cardboard sheet and fold up the sides to make a box. Cut too little or too much and the box is small; somewhere in between it holds the most. Derivatives find that sweet-spot cut precisely.
Where the slope is zero
At a maximum or minimum of a smooth function, the tangent line is flat — the derivative equals zero. Points where f′(x) = 0 are called critical points; checking them (and the endpoints) is how calculus finds the largest or smallest value.
If f′ changes from positive to negative at a critical point, it's a maximum; from negative to positive, a minimum. The sign of the slope tells you which.
For f(x) = −x² + 4x, f′(x) = −2x + 4 = 0 gives x = 2 — the peak, where f(2) = 4.
y = 1x² · slope at x=1 (the derivative 2a) = 2
Move along the parabola and watch the tangent's slope hit zero at the vertex.
- A rectangle's area is A(x) = x(10 − x).
- Differentiate and set A′(x) = 0.
- Solve for x to find the area-maximizing width.
What you should see: A′(x) = 10 − 2x = 0 gives x = 5, so a 5×5 square maximizes the area at 25.