7E - Optimisation
When tackling a maxima/minima question, use the following steps to help guide you through the problem:
- Wherever possible, draw a diagram and label all known information.
- Try to determine rules to represent the relationship between two variables. You may need to use known formula such as perimeter, area and volume. If appropriate, update you diagram with any new information.
- Identify the equation linking the variables to be maximised or minimised. Try to express in the form \(y=f(x)\).
- Find the derivative and then solve \(f'(x)=0\) for \(x\). This will provide information about stationary points.
- Verify that you have found the required maximum or minimum value. You can do this by sketching the graph of \(y=f(x)\) or by constructing a "sign" diagram (see Section 7C). At this point you should also check endpoints of the practical domain.
- Make sure you answer the question that has been asked. It is more than likely that you are asked for the maximum volume or area not just the value of \(x\) that results in this maximum or minimum point.
7E - Example 1: Applications of differentiation for optimisation (VCAA, 2006)
[VCAA, 2006 Exam 1 Question 9]
A rectangle \(WXYZ\) has two vertices, \(X\) and \(W\), on the \(x\)-axis and the other two vertices, \(Y\) and \(Z\), on the graph of \(y=9-3x^2\), as shown in the diagram below. The coordinates of \(Z\) are \((a,b)\) where \(a\) and \(b\) are positive real numbers.
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7E - Example 1: Video solution
7E - Example 1: Practice
7E - Example 1: Solutions
Question 1 [NZL, 2012]: The area of the triangle is given by \[A(x)=\frac{x}{2}\times (ax-x^2)\] To find maximum possible area solve \(A'(x)=0\) \[\therefore x=0, x=\frac{2a}{3}\] However, \(x>0\) \[\therefore A(\frac{2a}{3})=...=\frac{2a^3}{27}\] |
7E - Example 2: Applications of differentiation for optimisation (VCAA, 2008)
[VCAA, 2008 Exam 1 Question 9]
A plastic brick is made in the shape of a right triangular prism. The triangular end is an equilateral triangle with side length \(x\) cm and the length of the brick is \(y\) cm. The volume of the brick is \(1000\) \({cm}^3\).
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7E - Example 2: Video solution
7E - Example 2: Practice
Question 1: ABC Question 2: ABC 7E - Example 2: Solutions
Question 1: ABC Question 2: ABC |
7E - Example 3: Applications of differentiation for optimisation (VCAA, 2010)
[VCAA, 2010 Exam 1 Question 11]
A cylinder fits exactly in a right circular cone so that the base of the cone and one end of the cylinder are in the same plane as shown in the diagram below. The height of the cone is 5 cm and the radius of the cone is 2 cm. The radius of the cylinder is \(r\) cm and the height of the cylinder is \(h\) cm. For the cylinder inscribed in the cone as shown above
The total surface area, \(S\), of a cylinder of height \(h\) cm and radius \(r\) cm is given by \(S=2\pi rh+2\pi r^2\)
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7E - Example 3: Video solution
Video solution coming soon! (a) \(h=\frac{10-5r}{2}\) (b) \(S=10\pi r-3\pi r^2\) (c) \(S'(r)=0\therefore r=\frac{5}{3}\) 7E - Example 3: Practice
Question 1: ABC Question 2: ABC 7E - Example 3: Solutions
Question 1: ABC Question 2: ABC |