Economics at Zero Marginal Cost: Finding the Maximum and Minimum Value of a Function

1. Find the maximum and minimum value of the function $\\\\y=\frac{1}{3}x^{3}-3x^{2}+8x+25\\$

Solution:

Given: $\\\\y=\frac{1}{3}x^{3}-3x^{2}+8x+25\\$

To find: The maximum and minimum value of the function

First derivative:

$\\\\\frac{dy}{dx}=\frac{d}{dx} (\frac{1}{3}x^{3}-3x^{2}+8x+25)\\ \frac{dy}{dx}= \frac{d}{dx}\frac{1}{3}x^{3}-\frac{d}{dx}3x^{2}+\frac{d}{dx}8x+\frac{d}{dx}25\\ \frac{dy}{dx}= \frac{1}{3}\frac{d}{dx}x^{3}-3\frac{d}{dx}x^{2}+8\frac{d}{dx}x+0\\ \frac{dy}{dx}= \frac{1}{3}.3x^{3-1}-3.2x^{2-1}+8.1\\ \frac{dy}{dx}= x^{2}-6x+8\\$

First-order condition (Necessary condition):

$\\\frac{dy}{dx}= 0\ or\ x^{2}-6x+8=0\\$

Applying the quadratic formula, we have

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

$\\x^{2}-6x+8=0\\$

Here, a= 1, b= - 6, and c= 8

$\\x=\frac{-(-6))\pm \sqrt{(-6)^{2}-4.1.8}}{2.1}\ \\x=\frac{6\pm \sqrt{36-32}}{2}\ \\x=\frac{6\pm \sqrt{4}}{2}\ \\x=\frac{6\pm\2}{2}\ \\x=\frac{6+2}{2},\frac{6-2}{2}\ \\x=\frac{8}{2},\frac{4}{2}\\ \\x=4,2\$

∴ Stationary points are at x= 4 and x= 2

Second derivative:

$\\\\\frac{d^{2}y}{dx^{2}}=\frac{d}{dx} (x^{2}-6x+8)\\ \\\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}x^{2}-\frac{d}{dx}6x+\frac{d}{dx}8\\ \\\frac{d^{2}y}{dx^{2}}=2x^{2-1}-6\frac{d}{dx}x+0\\ \\\frac{d^{2}y}{dx^{2}}=2x-6.1\\ \\\frac{d^{2}y}{dx^{2}}=2x-6\\$

Second-order condition (Sufficient condition):

$\\\\At\ x=4\\ \\\frac{d^{2}y}{dx^{2}}=2x-6=2(4)-6=8-6=2>0\ \\\\At\ x=2\\ \\\frac{d^{2}y}{dx^{2}}=2x-6=2(2)-6=4-6=-2<0\\$

Hence, the function $\frac{1}{3}x^{3}-3x^{2}+8x+25\\$ is minimum at x=4 and maximum at x=2.

$\\y_{min}=\frac{1}{3}x^{3}-3x^{2}+8x+25\\ y_{min}=\frac{1}{3}(4)^{3}-3(4)^{2}+8(4)+25\\ y_{min}=\frac{1}{3}.64-3.16+32+25\\ y_{min}=\frac{64}{3}-48+57\\ y_{min}=\frac{64}{3}+9\\ y_{min}=\frac{91}{3}\\ y_{min}=30.33\\ \\y_{max}=\frac{1}{3}x^{3}-3x^{2}+8x+25\\ y_{max}=\frac{1}{3}(2)^{3}-3(2)^{2}+8(2)+25\\ y_{max}=\frac{1}{3}.8-3.4+16+25\\ y_{max}=\frac{8}{3}-12+41\\ y_{max}=\frac{8}{3}+29\\ y_{max}=\frac{95}{3}\\ y_{max}=31.67\\$

Therefore, the function $y=\frac{1}{3}x^{3}-3x^{2}+8x+25\\$ has a minimum value of $y_{min}=30.33\\$ and a maximum value of $y_{max}=31.67\\$ .

Plotting the function $y=\frac{1}{3}x^{3}-3x^{2}+8x+25\\$ on a graph:

2. Show that the curve $\\\\y=x+\frac{1}{x}\\$ has one maxima and one minima value.

Solution:

Given: $\\\\y=x+\frac{1}{x}\\$

To find: The maximum and minimum value of the function

First derivative:

$\\\\\frac{dy}{dx}=\frac{d}{dx}(x+\frac{1}{x})\ \\\frac{dy}{dx}=\frac{d}{dx}x+\frac{d}{dx}\frac{1}{x}\ \\\frac{dy}{dx}=1+\frac{d}{dx}x^{-1}\ \\\frac{dy}{dx}=1+(-1)x^{-1-1}\ \\\frac{dy}{dx}=1-x^{-2}\$

First-order condition (Necessary condition):

$\\\frac{dy}{dx}= 0\ or\ 1-x^{-2}=0\ or\ 1-\frac{1}{x^{2}}=0\\ or\ 1=\frac{1}{x^{2}}\\ or\ {x^{2}}=1\\ or\ x=\pm \sqrt1\\ or\ x=\pm 1\\ or\ x=1,-1\\$

∴ Stationary points are at x = 1 and x = −1

Second derivative:

$\\\\\frac{d^{2}y}{dx^{2}}=\frac{d}{dx} (1-x^{-2})\\ \\\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}1-\frac{d}{dx}x^{-2}\\ \\\frac{d^{2}y}{dx^{2}}=0-(-2)x^{-2-1}\\ \\\frac{d^{2}y}{dx^{2}}=2x^{-3}\\ \\\frac{d^{2}y}{dx^{2}}=\frac{2}{x^{3}}\\$

Second-order condition (Sufficient condition):

$\\\\At\ x=1\\ \\\frac{d^{2}y}{dx^{2}}=\frac{2}{x^{3}}=\frac{2}{(1)^{3}}=\frac{2}{1}=2>0 \\\\At\ x=-1\\ \\\frac{d^{2}y}{dx^{2}}=\frac{2}{x^{3}}=\frac{2}{(-1)^{3}}=\frac{2}{-1}=-2<0$

Hence, the function $\\\\y=x+\frac{1}{x}\\$ is maximum at x=−1 and minimum at x=1.

$\\y_{min}=x+\frac{1}{x}\\\\ y_{min}=1+\frac{1}{1}\\\ y_{min}=1+1\\\ y_{min}=2\\ \\y_{max}=x+\frac{1}{x}\\\\ y_{max}=(-1)+\frac{1}{(-1)}\\\ y_{max}=-1-1\\\ y_{max}=-2\\$