Maximum Values (Leaving Cert Mathematics): Revision Notes

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Maximum Values

What are maximum values?

In calculus, maximum values occur at the highest points on a curve or graph. These points are called turning points because the direction of the curve changes from increasing to decreasing at a maximum.

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The key principle to remember is that at any maximum or minimum turning point, the derivative equals zero:

dydx=0\frac{dy}{dx} = 0

This fundamental rule allows us to find maximum values by differentiating a function and setting the result equal to zero.

Figure

The diagram above shows a typical maximum point on a parabola where the derivative equals zero at the peak.

Why are maximum values important?

Maximum value problems appear frequently in real-life situations where we need to optimise something - finding the best possible result given certain constraints.

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Common applications include:

  • Maximising profit in business situations
  • Maximising area with limited materials
  • Finding peak performance in engineering
  • Optimising trajectories in physics

The beauty of calculus is that it gives us a systematic method to solve these problems rather than just guessing.

Step-by-step method for finding maximum values

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Method for Finding Maximum Values:

  1. Write down the function you want to maximise
  2. Differentiate the function with respect to the variable
  3. Set the derivative equal to zero and solve for the variable
  4. Substitute back into the original function to find the maximum value
  5. Check your answer makes sense in the context

Worked example: projectile motion

Let's examine a rocket fired into the air. The height h (in metres) after t seconds is given by:

h(t)=−2t2+28th(t) = -2t^2 + 28t

Figure

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Worked Example: Projectile Motion

Part (i): How long is the rocket in flight?

The rocket is in flight between when it leaves the ground (h = 0) and when it lands (h = 0 again).

Setting h = 0: −2t2+28t=0-2t^2 + 28t = 0 2t(−t+14)=02t(-t + 14) = 0

This gives us t = 0 (launch) or t = 14 (landing).

Therefore, the rocket is in flight for 14 seconds.

Part (ii): What is the maximum height?

To find the maximum height, we use our method:

  1. The function is: h=−2t2+28th = -2t^2 + 28t
  2. Differentiate: dhdt=−4t+28\frac{dh}{dt} = -4t + 28
  3. Set equal to zero: −4t+28=0-4t + 28 = 0 Solving: 4t=284t = 28, so t=7t = 7
  4. Substitute t = 7 back into the original equation: h=−2(7)2+28(7)=−98+196=98h = -2(7)^2 + 28(7) = -98 + 196 = 98

The maximum height reached is 98 metres at t = 7 seconds.

Area optimisation problems

A common type of maximum value problem involves maximising area with constraints on perimeter or available materials.

Figure

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Worked Example: Area Optimisation

A farmer has 20 metres of fencing and wants to create a rectangular enclosure in the corner of a field (using the field boundaries for two sides).

If the width is x metres, then the length is (20 - x) metres.

The area is: A=x(20−x)=20x−x2A = x(20 - x) = 20x - x^2

To find maximum area:

  1. Differentiate: dAdx=20−2x\frac{dA}{dx} = 20 - 2x
  2. Set equal to zero: 20−2x=020 - 2x = 0, so x=10x = 10
  3. Maximum area = 10×10=10010 \times 10 = 100 square metres

Common exam question types

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Typical Maximum Value Problems:

  • Projectile motion: Objects thrown or fired follow parabolic paths. Find when they reach maximum height and what that height is.
  • Area problems: Given constraints on perimeter or materials, find dimensions that maximise area.
  • Business optimisation: Find production levels that maximise profit or minimise cost.
  • Geometric problems: Optimise dimensions of shapes like rectangles, triangles, or cylinders.

Exam tips

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Essential Exam Strategies:

  • Always check your derivative calculation - this is where most errors occur
  • Remember the context - negative values might not make physical sense
  • Show all steps clearly - partial marks are awarded for method
  • Substitute back to find the actual maximum value, not just where it occurs
  • Use appropriate units in your final answer

Identifying maximum vs minimum

Understanding the shape of curves helps you identify whether you have found a maximum or minimum point.

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Visual Identification:

On a graph:

  • Maximum: the curve goes up then down (∩ shape)
  • Minimum: the curve goes down then up (∪ shape)

For quadratic functions:

  • Negative coefficient of x²: gives a maximum (opens downward)
  • Positive coefficient of x²: gives a minimum (opens upward)

Figure

This parabola opens downward, so it has a maximum point at its vertex.

Remember!

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Key Points to Remember:

  • The derivative equals zero at all maximum and minimum points: dydx=0\frac{dy}{dx} = 0
  • Follow the systematic method: differentiate, set equal to zero, solve, then substitute back
  • Maximum value problems appear in many real-life contexts - projectile motion, business optimisation, and area problems are the most common
  • Check your answer makes sense in the context of the problem - negative times or impossible dimensions indicate an error
  • Practice identifying whether you have a maximum or minimum by looking at the shape of the curve or the sign of the coefficient

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