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Calculus - Optimisation

Grade 11IB_AA

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

🔑Concepts

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Optimization is the process of finding the maximum or minimum value of a function, often subject to constraints.

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A stationary point occurs where the first derivative of the function is zero: f′(x)=0f'(x) = 0.

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To verify the nature of a stationary point, use the second derivative test: if f′′(x)>0f''(x) > 0, the point is a local minimum; if f′′(x)<0f''(x) < 0, it is a local maximum.

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In real-world problems, you must often use a constraint equation (e.g., fixed volume or perimeter) to substitute one variable and express the objective function in terms of a single variable.

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Always check the endpoints of the domain if the function is defined on a closed interval [a,b][a, b], as the absolute maximum or minimum might occur there.

📐Formulae

dydx=0(Condition for stationary points)\frac{dy}{dx} = 0 \quad \text{(Condition for stationary points)}

d2ydx2<0  ⟹  Local Maximum\frac{d^2y}{dx^2} < 0 \implies \text{Local Maximum}

d2ydx2>0  ⟹  Local Minimum\frac{d^2y}{dx^2} > 0 \implies \text{Local Minimum}

Arectangle=l×wA_{rectangle} = l \times w

Vcylinder=πr2hV_{cylinder} = \pi r^2 h

Scylinder=2πr2+2πrhS_{cylinder} = 2\pi r^2 + 2\pi rh

💡Examples

Problem 1:

A closed rectangular box has a square base of side length xx cm and a height of hh cm. The total surface area of the box is 600 cm2600 \text{ cm}^2. Find the maximum volume of the box.

Solution:

  1. Express the surface area: A=2x2+4xh=600A = 2x^2 + 4xh = 600.
  2. Solve for hh: 4xh=600−2x2  ⟹  h=600−2x24x=150x−x24xh = 600 - 2x^2 \implies h = \frac{600 - 2x^2}{4x} = \frac{150}{x} - \frac{x}{2}.
  3. Express Volume VV: V=x2h=x2(150x−x2)=150x−12x3V = x^2 h = x^2 \left( \frac{150}{x} - \frac{x}{2} \right) = 150x - \frac{1}{2}x^3.
  4. Find the derivative: dVdx=150−32x2\frac{dV}{dx} = 150 - \frac{3}{2}x^2.
  5. Set dVdx=0\frac{dV}{dx} = 0: 150=32x2  ⟹  x2=100  ⟹  x=10150 = \frac{3}{2}x^2 \implies x^2 = 100 \implies x = 10 (since x>0x > 0).
  6. Check second derivative: d2Vdx2=−3x\frac{d^2V}{dx^2} = -3x. For x=10x = 10, d2Vdx2=−30\frac{d^2V}{dx^2} = -30, which is <0< 0, confirming a maximum.
  7. Calculate Max Volume: V=150(10)−12(10)3=1500−500=1000 cm3V = 150(10) - \frac{1}{2}(10)^3 = 1500 - 500 = 1000 \text{ cm}^3.

Explanation:

First, we use the surface area constraint to eliminate hh. Then, we differentiate the volume function and solve for the critical value of xx. Finally, we verify it is a maximum using the second derivative test.

Problem 2:

A farmer wants to enclose a rectangular paddock using an existing straight stone wall as one side. He has 200200 m of fencing for the other three sides. Find the dimensions that provide the maximum area.

Solution:

  1. Let xx be the width perpendicular to the wall and yy be the length parallel to the wall.
  2. Constraint: 2x+y=200  ⟹  y=200−2x2x + y = 200 \implies y = 200 - 2x.
  3. Area A=xy=x(200−2x)=200x−2x2A = xy = x(200 - 2x) = 200x - 2x^2.
  4. Differentiate: dAdx=200−4x\frac{dA}{dx} = 200 - 4x.
  5. Set to zero: 200−4x=0  ⟹  x=50200 - 4x = 0 \implies x = 50.
  6. Find yy: y=200−2(50)=100y = 200 - 2(50) = 100.
  7. Dimensions are 50 m50 \text{ m} by 100 m100 \text{ m}.

Explanation:

Since one side is a wall, the fencing only covers three sides (2x+y2x + y). We substitute the constraint into the area formula to create a quadratic function and find its vertex/maximum.