Answers

  1. \(y = 2 + 6x\)

\(\frac{d(y)}{d(x)} = 6\)

  1. \(y = 5 - 4x +2x^3\)

\(\frac{d(y)}{d(x)} = -4 + 6x^2\)

  1. \(y = 25 +6x^2 - 3x^3 +25x^4\)

\(\frac{d(y)}{d(x)} = 12x - 9x^2 + 100x^3\)

  1. \(y - 3 = 2x\)

\(\frac{d(y)}{d(x)} = 2\)

  1. \(TPP = 24 +5L +2L^2 - L^3\)

\(\frac{d(TPP)}{d(L)} = 5 +4L - 3L^2\)

Find the second derivative of the following

  1. \(TU = 25 + X_1 - X_1^2\)

\(\frac{dy}{dx} = 1 - 2X_1\)

\(\frac{d^2(U)}{d(X_1)} = -2\)

  1. \(TU = 25 +25X_1 -2X_1^2\)

\(\frac{dy}{dx} = 25 - 4X_1\)

\(\frac{d^2(U)}{d(X_1)} = - 4\)

There is diminishing marginal utility. The first derivative is the slope of the utility relative to the consumption of product \(X_1\). This is marginl utility. The second derivative shows how the slope or marginal utility changes with \(X_1\). The result is -2 or -4. Therefore, the slope is falling as the consumption of \(X_1\) increases. This is diminishing marginal utility.

  1. \(TPP = 15 +15Q +Q^2 - Q^3\)

\(\frac{TPP}{dQ} = 15 + 2Q - 3Q^2\)

\(\frac{d^2(TPP)}{d(Q)} = 2 -6Q\)

  1. What does your answer to the previous question tell you about the shape of the Total Physical Product Curve?

There are deminishing returns. The first derivative is the change in output as more input (Q) is added. This slope is the marginal return to adding an input. It is \(15 + 2Q - 3Q^2\) meaning that as more (say) labour is added, output rises at a faster rate and then starts to rise at slower rate. The second derivtive shows how the rate of change in output changes over time. This is \(2 - 6Q\). Therefore it is positive or a short while (when \(Q > \frac{1}{3}\)) and then turns negative. In other words, it accelerates and then decelerates. These are the diminishing marginal returns.

  1. Given a demand curve

\[Q_d = 90 - 5P + 0.2P^2\]

What is the elasticity of demand at the point \(P = 5, Q = 70\)?

The equation for elasticity is

\[\varepsilon_d = \frac{d(Q)}{d(P)}\times \frac{P}{Q}\]

\[\frac{dQ}{dP} = -5 + 0.4P\]

Therefore,

\[\varepsilon_d = \frac{-5 + (0.4 \times 5)}\times \frac{5}{70}\] \[\varepsilon_d = -5 + 2 \times \frac{5}{70}\] \[\varepsilon_d = - \frac{15}{70}\] \[\varepsilon_d = -\frac{3}{14}\]

As the absolute value is below one, the demand is inelastic

  1. Here is the total physical product curve. Show the average physical product and the marginal physical product at the point where 8 units of labour are being used.

The marginal product is tangent to the red line at the point where L = 8.

  1. Given the \(TPP = 500 + 180Q + 15Q^2 - 2Q^3\),

Find equation for MPP

MPP is the marginal physical product or the slope of the total physical product

\[\frac{d(TPP)}{d(Q)} = 18 + 30Q - 6Q^2\]

The gradient of the TPP is zero at its peak.

This cannot be solved by factorising. You will have to use the equation

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

Input this into your calculator or click the following link.

On line quadratic solver

Remember that, a, b and c are given by the equation

\[ax^2 + bx + c\]

  1. Given the \(TPP = 100 + 32Q +23Q^2 - Q^3\),

Find the MPP

\[MPP = 32 + 46Q - 3Q^2\]

Set MPP equal to zero

\[3Q^2 - 46Q -32 = 0\] \[(3Q + 2)(Q - 16) = 0\]

Therefore \[3Q = -2 \quad \text{or} \quad Q = 16\]

\(Q = 16\) at the maximum.