Find and classify the stationary points of y = x³ − 3x² − 9x + 5, and find the point of inflection

1. y′ = 3x² − 6x − 9 = 3(x² − 2x − 3) = 3(x − 3)(x + 1), so y′ = 0 at x = 3 and x = −1.
2. y″ = 6x − 6. At x = −1: y″ = −12 < 0, maximum. At x = 3: y″ = 12 > 0, minimum.
3. y(−1) = −1 − 3 + 9 + 5 = 10 and y(3) = 27 − 27 − 27 + 5 = −22.
4. Maximum at (−1, 10), minimum at (3, −22).
5. Inflection: y″ = 0 at x = 1, and y″ changes from negative to positive there. y(1) = 1 − 3 − 9 + 5 = −6, so the point of inflection is (1, −6).

When the second-derivative test fails: y = x⁴

1. y′ = 4x³ = 0 at x = 0. y″ = 12x², and y″(0) = 0, so the test is inconclusive.
2. Sign table for y′: at x = −1, y′ = −4 (negative); at x = 1, y′ = 4 (positive).
3. The slope changes from − to +, so (0, 0) is a minimum, even though y″ = 0 there.

A farmer has 100 m of fencing for a rectangular field against a straight river. No fence is needed along the river. Find the maximum area.

1. Let the two sides perpendicular to the river be x m and the side parallel be y m. Fence used: 2x + y = 100, so y = 100 − 2x (with 0 < x < 50).
2. Area: A = xy = x(100 − 2x) = 100x − 2x²
3. A′ = 100 − 4x = 0 gives x = 25. A″ = −4 < 0, so this is a maximum.
4. y = 100 − 50 = 50. Maximum area = 25 × 50 = 1250 m[sup 2].

Squares of side x cm are cut from the corners of a 24 cm × 24 cm sheet, which is folded into an open box. Find the x that gives the largest volume.

1. V = x(24 − 2x)², for 0 < x < 12.
2. Product rule: V′ = (24 − 2x)² + x × 2(24 − 2x)(−2) = (24 − 2x)[(24 − 2x) − 4x] = (24 − 2x)(24 − 6x)
3. V′ = 0 gives x = 12 (volume zero, reject) or x = 4.
4. Sign check: V′(3) = 18 × 6 > 0 and V′(5) = 14 × (−6) < 0, so x = 4 is a maximum.
5. Maximum volume: V(4) = 4 × 16² = 1024 cm[sup 3] when x = 4 cm.

A closed cylindrical can must hold 500 cm³. Find the radius that minimizes the surface area.

1. Constraint: πr²h = 500, so h = 500πr².
2. Surface area: S = 2πr² + 2πrh = 2πr² + 2πr × 500πr² = 2πr² + 1000r
3. S′ = 4πr − 1000r² = 0, so 4πr³ = 1000 and r³ = 250π = 79.58…, giving r = 4.30 cm (to 2 decimal places).
4. S″ = 4π + 2000r³ > 0 for all r > 0, so this is a minimum.
5. r ≈ 4.30 cm (and then h = 500πr² ≈ 8.60 cm, which is 2r — the most efficient can is as tall as it is wide).