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Calculus - Stationary Points (Max/Min)

Grade 12IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A stationary point occurs on a curve where the gradient is zero, i.e., dy/dx = 0.

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There are three main types of stationary points: Local Maximum, Local Minimum, and Stationary Point of Inflexion.

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The Second Derivative Test is the most common method to determine the nature of a stationary point.

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A function is increasing when dy/dx > 0 and decreasing when dy/dx < 0.

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If the second derivative is zero, the nature must be checked by testing the gradient slightly to the left and right of the point (First Derivative Test).

📐Formulae

Stationary Point Condition: dydx=0\frac{dy}{dx} = 0

Local Minimum: d2ydx2>0\frac{d^2y}{dx^2} > 0 at the stationary point

Local Maximum: d2ydx2<0\frac{d^2y}{dx^2} < 0 at the stationary point

Point of Inflexion (Possible): d2ydx2=0\frac{d^2y}{dx^2} = 0

💡Examples

Problem 1:

Find the coordinates and the nature of the stationary points for the curve y=x3−3x+2y = x^3 - 3x + 2.

Solution:

  1. Find dydx\frac{dy}{dx}: 3x2−33x^2 - 3.
  2. Set dydx=0\frac{dy}{dx} = 0: 3(x2−1)=0  ⟹  x=1,x=−13(x^2 - 1) = 0 \implies x = 1, x = -1.
  3. Find yy-coordinates: At x=1,y=(1)3−3(1)+2=0x = 1, y = (1)^3 - 3(1) + 2 = 0. At x=−1,y=(−1)3−3(−1)+2=4x = -1, y = (-1)^3 - 3(-1) + 2 = 4.
  4. Find d2ydx2\frac{d^2y}{dx^2}: 6x6x.
  5. Test x=1x = 1: d2ydx2=6(1)=6>0\frac{d^2y}{dx^2} = 6(1) = 6 > 0 (Minimum).
  6. Test x=−1x = -1: d2ydx2=6(−1)=−6<0\frac{d^2y}{dx^2} = 6(-1) = -6 < 0 (Maximum). Final Answer: (1, 0) is a Local Minimum; (-1, 4) is a Local Maximum.

Explanation:

We first differentiated the function to find the x-values where the slope is zero. After finding the corresponding y-coordinates, we used the second derivative to check the concavity: a positive second derivative indicates a 'cup' shape (minimum) and a negative one indicates a 'cap' shape (maximum).

Problem 2:

A closed rectangular box has a square base of side xx cm and a total surface area of 150150 cm². Express the height hh in terms of xx and find the value of xx that maximizes the volume.

Solution:

  1. Surface Area: 2x2+4xh=150  ⟹  4xh=150−2x2  ⟹  h=150−2x24x2x^2 + 4xh = 150 \implies 4xh = 150 - 2x^2 \implies h = \frac{150 - 2x^2}{4x}.
  2. Volume V=x2h=x2(150−2x24x)=150x−2x34=37.5x−0.5x3V = x^2h = x^2(\frac{150 - 2x^2}{4x}) = \frac{150x - 2x^3}{4} = 37.5x - 0.5x^3.
  3. Maximize: dVdx=37.5−1.5x2\frac{dV}{dx} = 37.5 - 1.5x^2.
  4. Set dVdx=0:1.5x2=37.5  ⟹  x2=25  ⟹  x=5\frac{dV}{dx} = 0: 1.5x^2 = 37.5 \implies x^2 = 25 \implies x = 5 (since x>0x > 0).
  5. Verify nature: d2Vdx2=−3x\frac{d^2V}{dx^2} = -3x. At x=5x=5, −15<0-15 < 0, so it is a maximum.

Explanation:

In optimization problems, we express the target variable (Volume) in terms of a single variable (x) using the given constraint (Surface Area). We then differentiate and solve for zero to find the stationary point, verifying it is a maximum using the second derivative.