Quantitative methods: basic mathematics

Mathematics is used in economics to express economic relationships as equations (e.g. demand and supply functions) and to calculate and compare economic quantities such as cost, revenue and elasticity.

The differential coefficient of $3x^2$ is $6x$.

$\frac{d}{dx}(3x^2)=3\cdot 2x^{2-1}=6x$

Slope $m=\dfrac{y_2-y_1}{x_2-x_1}$

$m=\frac{7-3}{5-2}=\frac{4}{3}$

So the slope of the line AB is $\dfrac{4}{3}$.

Let the line make equal intercepts on both axes.
Let each intercept be $a$.

Equation of the line in intercept form is:

$\frac{x}{a} + \frac{y}{a} = 1$

Since the line passes through (A(-5, 10)),

$\frac{-5}{a} + \frac{10}{a} = 1$

$\frac{5}{a} = 1$

$a = 5$

Substituting (a = 5) in the intercept form:

$\frac{x}{5} + \frac{y}{5} = 1$

Multiplying both sides by 5:

$x + y = 5$

Hence, the required equation of the straight line is:

$\boxed{x + y = 5} \quad \text{or} \quad \boxed{x + y - 5 = 0}$

The line makes intercepts of 5 units on both the x-axis and y-axis.

Note: The equation did not print in the question. The standard form of this problem is taken as $Q = 20L - L^2$; the method below works for whatever function is given — marginal product is simply the first derivative of the total product function with respect to labour.

Given: Total Product function: $Q = 20L - L^2$ where $Q$ = total output and $L$ = units of labour.

a. Marginal production function

Marginal product of labour ($MP_L$) is the addition to total product from employing one more unit of labour. It is obtained by differentiating the total product function with respect to labour:

$MP_L = \frac{dQ}{dL} = \frac{d}{dL}\left(20L - L^2\right)$

$MP_L = 20 - 2L$

So the marginal production function is $MP_L = 20 - 2L$.

b. Marginal product when labour = 5 units

Put $L = 5$ in the marginal production function:

$MP_L = 20 - 2(5)$ $MP_L = 20 - 10$ $MP_L = 10 \text{ units}$

Answer: When 5 units of labour are employed, the marginal product of labour is 10 units of output — that is, the 5th unit of labour adds 10 units to total output.

Let the $Y$-intercept be $b$. Then the $X$-intercept is $3b$, and the equation in intercept form is $\frac{x}{3b} + \frac{y}{b} = 1 \Rightarrow x + 3y = 3b$

The line passes through $(6, 4)$, so $6 + 3(4) = 3b \Rightarrow 18 = 3b \Rightarrow b = 6$

Hence $3b = 18$ and the required equation is $x + 3y = 18$

Check: when $y = 0$, $x = 18$; when $x = 0$, $y = 6$. The $X$-intercept $18 = 3 \times 6$, and $(6, 4)$ gives $6 + 12 = 18$.

Given $4x + 6y = 18$. Writing it in the form $y = mx + c$: $6y = -4x + 18 \Rightarrow y = -\frac{2}{3}x + 3$

Comparing with $y = mx + c$:

• Slope $m = -\dfrac{2}{3}$
• $Y$-intercept $c = 3$

Check: at $x = 0$, $6y = 18$, so $y = 3$.

Given $y = 5x^3 - 4x + 7$.

$\frac{dy}{dx} = 5 \times 3x^2 - 4 + 0 = 15x^2 - 4$

Slope $m = \dfrac{y_2 - y_1}{x_2 - x_1}$, with $(x_1, y_1) = (3, 4)$ and $(x_2, y_2) = (7, 12)$: $m = \frac{12 - 4}{7 - 3} = \frac{8}{4} = 2$ The slope is 2.

Let the line have equal intercepts $a$ on both axes. Its equation is $\frac{x}{a} + \frac{y}{a} = 1 \Rightarrow x + y = a$

It passes through $(4, 6)$, so $4 + 6 = a \Rightarrow a = 10$

Hence the required equation is $x + y = 10$

Check: the intercepts are $x = 10$ (when $y = 0$) and $y = 10$ (when $x = 0$), which are equal, and $(4, 6)$ satisfies $4 + 6 = 10$.

Given $TC = 100 + 15Q - 6Q^2 + Q^3$.

Average cost $AC = \frac{TC}{Q} = \frac{100}{Q} + 15 - 6Q + Q^2$

Marginal cost $MC = \frac{d(TC)}{dQ} = 15 - 12Q + 3Q^2$

Here 100 is the fixed cost, and its derivative is zero, so it does not affect MC.

A straight line is the path traced by a point that moves in a constant direction. Algebraically, it is the graph of a linear equation in two variables of the form $y = mx + c$ (or $ax + by + c = 0$), where $m$ is the slope and $c$ is the $y$-intercept.

Given: $A(2,4)$ and $B(6,12)$.

Slope

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{12-4}{6-2}=\frac{8}{4}=2$

Equation (point-slope form)

$y-y_1=m(x-x_1)$

$y-4=2(x-2)$

$y-4=2x-4$

$y=2x$

Check: At $B$, $x=6$ gives $y=2(6)=12$, which is correct.

Hence the equation of the line is $y=2x$ (or $2x-y=0$) and its slope is $2$.

Given: $Q=100L-2L^2$

The marginal product is the first derivative of total output with respect to labour:

$MP=\frac{dQ}{dL}=100-4L$

At $L=5$:

$MP=100-4(5)=100-20=80$

Hence $MP=100-4L$, and at $L=5$ the marginal product is 80 units.

The standard form of a straight-line equation is $y = mx + c$, where $m$ is the slope and $c$ is the intercept on the $y$-axis. (It may also be written as $ax + by + c = 0$.)

Using the power rule, $\frac{d}{dx}(x^n) = nx^{n-1}$:

$\frac{dy}{dx} = 5(2x) + 3 = 10x + 3$

Marginal cost is the derivative of total cost with respect to output:

$MC = \frac{d(TC)}{dQ}$

Given $TC = Q^3 - 6Q^2 + 15Q + 100$,

$MC = 3Q^2 - 12Q + 15$

The constant 100 is fixed cost, so its derivative is zero. Hence $MC = 3Q^2 - 12Q + 15$.