Applications of differential calculus

Differential calculus plays a crucial role in analysing changes and optimising solutions across numerous fields. The mathematical aspects of differential calculus that are used in everyday life involve understanding rates of change, optimisation, and the behaviour of functions.

Some considerations and applications of differential calculus are:

  • Derivative as a rate of change
  • Tangent lines and slopes
  • Optimisation
  • Higher order derivatives
  • Limits and continuity

These mathematical tools from differential calculus form the foundation for analysing, modelling, and optimising systems in a variety of fields. Their application is essential for solving complex problems that involve change and variability.

Worked Example

Example 1

A company produces and sell generators. The revenue \(R\left(x\right)\) and cost \(C\left(x\right)\) are in dollars, where \(x\) is the number of units produced and sold. The profit can be modeled by

\[P\left(x\right) = R\left(x\right) - C\left(x\right)\]

The cost to produce one unit is \(\$800\) i.e. \(R\left(x\right) = 800x\). the cost function is modeled by

\[C\left(x\right) = 10x^{2} + 40x +2000\]

Write the profit function, and from it, determine the number of units that should be produced and sold to maximise the profit.

Solution

\[\begin{align}P(x) &= R(x) - C(x) \\ P(x) &= 800x - (10x^{2} + 40x + 2000)\end{align}\]

A graph of the profit function is

A graph of the quadratic function f(x) = -10x^2 + 760x + 2000. The parabola opens downward, indicating a maximum point. The turning point, which represents the maximum, is labelled as (38, 12440). The graph exhibits typical quadratic behaviour, increasing to the maximum and then decreasing symmetrically.

From the graph it can be seen:

  1. Profit is made only when the graph is above the \(x\)-axis i.e when \(3 \le\ x\ \le73;\ x \in z\)
  2. 2.The maximum profit is $12440.00 when 38 generators are sold (the turning point)

Calculus

\[\begin{align} &P(x)=-10x^2+760x-2000\\
&P^\prime(x) = -20x + 760 \\
&P^\prime(x) = 0 \Rightarrow -20x + 760 = 0 \\ & \quad\quad \Rightarrow x = 38 \text{ units} \\
&P(38) = -10(38)^2 + 760(38) - 2000 = 12440 \quad \text{(Max profit)} \\
&P^\prime(37) > 0 \quad \text{and} \quad P'(39) < 0 \therefore \text{Local maximum} \\
&P^{\prime\prime}(x) = -20 \Rightarrow P^{\prime\prime}(38) < 0 \therefore \text{Local maximum}
\end{align} \]

Example 2

Consider an open rectangular prism (box) formed from a thin metal sheet. To form the box, square sections are cut out from each of the 4 corners, and the sheet is folded at right angles in the upward direction in line with where the squares are cut out. Determine the maximum volume of the box so formed. Determine the size of the square edge cut out and the maximum volume of the box so formed.

The longer edge of the rectangular sheet is 20 units and the shorter edge is 12 units.

A diagram illustrating the process of creating an open-top rectangular box from a 12 cm by 20 cm rectangular sheet. On the left, the flat sheet is shown with four identical squares of side length x cut from each corner. The remaining flaps are folded up to form the sides of the box. On the right, the resulting three-dimensional open-top box is depicted. The box's height is x, while its base dimensions are (20 - 2x) cm by (12 - 2x) cm.

Solution

\(\begin{align} &l = 20 -2x \\ &w = 12 - 2x \\ &h = x \end{align}\)

\(\begin{align}&V\left(l, w, h\right) = \left(20 – 2x\right) \cdot\left(12 – 2x\right)\cdot x; 0 < x < 6 \\ &V\left(x\right) = \left(20 – 2x\right)\cdot \left(12 – 2x\right) \cdot x \\ &V\left(x\right) = 4x^{3} – 64x^{2} + 240x \end{align}\)

\(\begin{align} V^\prime\left(x\right) = 0 &\Rightarrow 12x^{2}-128x+240=0 \\ &\Rightarrow x = 2.427\ \text{or}\ x = 8.239 \end{align}\)

\(\begin{align}&x = 2.427\ \text{as}\ x = 8.239\ \text{is not in the domain} \\ &V\left(2.427\right) = 262.682\ \text{cubic units} \\ &\left(2.427, 262.682\right)\ \text{is a local maximum}\end{align}\)

A graph of the function f(x) = 4x^3 - 64x^2 + 240x for the interval 0 < x < 6. The curve has a local maximum at (2.43, 262.68). The function starts from a positive value at x = 0, increases to the maximum, and then decreases towards x = 6. The graph exhibits typical cubic behaviour within the given domain, forming a peak before declining.