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

1. Find the maximum and minimum value of the function $\\\\y=3x^3+2x^{2}-6x+6\\$

Solution:

Given: $\\\\y=3x^3+2x^{2}-6x+6\\$

To find: The maximum and minimum value of the function.

First derivative:

$\\\\\frac{dy}{dx}=\frac{d}{dx}(3x^3+2x^{2}-6x+6)\ \\\frac{dy}{dx}=\frac{d}{dx}3x^3+\frac{d}{dx}2x^{2}-\frac{d}{dx}6x+\frac{d}{dx}6\ \\\frac{dy}{dx}=3\frac{d}{dx}x^3+2\frac{d}{dx}x^{2}-6\frac{d}{dx}x+0\ \\\frac{dy}{dx}=3.3x^{3-1}+2.2x^{2-1}-6.1\ \\\frac{dy}{dx}=9x^{2}+4x-6\$

First-order condition (Necessary condition):

$\\\frac{dy}{dx}= 0\ or\ 9x^{2}+4x-6=0\\$

Applying the quadratic formula, we have

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

$9x^{2}+4x-6=0\\$

Here, a= 9, b= 4, and c= - 6

$\\x=\frac{-4\pm \sqrt{(4)^{2}-4.9.-6}}{2.9}\\ \\x=\frac{-4\pm \sqrt{16+216}}{18}\\ \\x=\frac{-4\pm \sqrt{232}}{18}\\ \\x=\frac{-4\pm \15.23}{18}\\ \\x=\frac{-4+15.23}{18},\frac{-4-15.23}{18}\\ \\x=\frac{11.23}{18},\frac{19.23}{18}\\ \\x=0.624,-1.068\\$

∴ Stationary points are at x= 0.624 and x= - 1.068.

Second derivative:

$\\\\\frac{d^{2}y}{dx^{2}}=\frac{d}{dx} (9x^{2}+4x-6)\\ \\\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}9x^{2}+\frac{d}{dx}4x-\frac{d}{dx}6\\ \\\frac{d^{2}y}{dx^{2}}=9\frac{d}{dx}x^{2}+4\frac{d}{dx}x-0\\ \\\frac{d^{2}y}{dx^{2}}=9.2x^{2-1}+4.1\\ \\\frac{d^{2}y}{dx^{2}}=18x+4\\$

Second-order condition (Sufficient condition):

$\\\\At\ x=0.624\\ \\\frac{d^{2}y}{dx^{2}}=18x+4=18(0.624)+4=11.232+4=15.232>0\ \\\\At\ x=-1.068\\ \\\frac{d^{2}y}{dx^{2}}=18x+4=18(-1.068)+4=-19.224+4=-15.224<0$

Hence, the function $\\\\y=3x^3+2x^{2}-6x+6\\$ is minimum at x=0.624 and maximum at x=-1.068.

$\\y_{min}=3x^3+2x^{2}-6x+6\\ y_{min}=3(0.624)^3+2(0.624)^{2}-6(0.624)+6\\ y_{min}=3(0.243)+2(0.389)-3.744+6\\ y_{min}=0.729+0.778-3.744+6\\ y_{min}=3.763\\ \\y_{max}=3x^3+2x^{2}-6x+6\\ y_{max}=3(-1.068)^3+2(-1.068)^{2}-6(-1.068)+6\\ y_{max}=3(-1.218)+2(1.141)+6.408+6\\ y_{max}=-3.654+2.282+6.408+6\\ y_{max}=11.036\\$

Therefore, the function $\\\\y=3x^3+2x^{2}-6x+6\\$ has a maxima value of $y_{max}=11.036\\$ and a minima value of $y_{min}=3.763\\$ .

Plotting the function $\\\\y=3x^3+2x^{2}-6x+6\\$ on a graph:

2. For which value the function $\\\\y=x^3-12x^{2}+36x+8\\$ has a maximum or minimum value?

Solution:

Given: $\\\\y=x^3-12x^{2}+36x+8\\$

To find: The value of x at which the function has a maximum or minimum value.

First derivative:

$\\\\\frac{dy}{dx}=\frac{d}{dx}(x^3-12x^{2}+36x+8)\ \\\frac{dy}{dx}=\frac{d}{dx}x^3-\frac{d}{dx}12x^{2}+\frac{d}{dx}36x+\frac{d}{dx}8\ \\\frac{dy}{dx}=3x^{3-1}-12\frac{d}{dx}x^{2}+36\frac{d}{dx}x+0\ \\\frac{dy}{dx}=3x^{2}-12.2x^{2-1}+36.1\ \\\frac{dy}{dx}=3x^{2}-24x+36\$

First-order condition (Necessary condition):

$\\\frac{dy}{dx}= 0\ or\3x^{2}-24x+36=0\$

Applying the quadratic formula, we have

$\\x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\\\ 3x^{2}-24x+36=0\\$

Here, a= 3, b= -24, and c= 36

$\\x=\frac{-(-24)\pm \sqrt{(-24)^{2}-4.3.36}}{2.3}\\ \\x=\frac{24\pm \sqrt{576-432}}{6}\\ \\x=\frac{24\pm \sqrt{144}}{6}\\ \\x=\frac{24\pm \12}{6}\\ \\x=\frac{24+12}{6},\frac{24-12}{6}\\ \\x=\frac{36}{6},\frac{12}{6}\\ \\x=6,2\\$

∴ Stationary points are at x= 6 and x= 2.

Second derivative:

$\\\\\frac{d^{2}y}{dx^{2}}=\frac{d}{dx} (3x^{2}-24x+36)\\ \\\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}3x^{2}-\frac{d}{dx}24x+\frac{d}{dx} 36\\ \\\frac{d^{2}y}{dx^{2}}=3\frac{d}{dx}x^{2}-24\frac{d}{dx}x+0\\ \\\frac{d^{2}y}{dx^{2}}=3.2x^{2-1}-24.1\\ \\\frac{d^{2}y}{dx^{2}}=6x-24\\$

Second-order condition (Sufficient condition):

$\\\\At\ x=6\\ \\\frac{d^{2}y}{dx^{2}}=6x-24=6(6)-24=36-24=12>0\ \\\\At\ x=2\\ \\\frac{d^{2}y}{dx^{2}}=6x-24=6(2)-24=12-24=-12<0$

Hence, the function $\\\\y=x^3-12x^{2}+36x+8\\$ is minimum at x=6 and maximum at x=2.

Plotting the function $\\\\y=x^3-12x^{2}+36x+8\\$ on a graph:

3. If $\\\\y=15-3x^{2}+x^{3}\\$ , find the value of x for which f(x) is maximum.

Solution:

Given: $\\\\y=15-3x^{2}+x^{3}\\$

To find: The value of x for which the function is maximum.

First derivative:

$\\\\\frac{dy}{dx}=\frac{d}{dx}(15-3x^{2}+x^{3})\ \\\frac{dy}{dx}=\frac{d}{dx}15-\frac{d}{dx}3x^{2}+\frac{d}{dx}x^{3}\ \\\frac{dy}{dx}=0-3\frac{d}{dx}x^{2}+3x^{3-1}\ \\\frac{dy}{dx}=-3.2x^{2-1}+3x^{2}\ \\\frac{dy}{dx}=-6x+3x^{2}\\\\ or\\ \\\frac{dy}{dx}=3x^{2}-6x\$

First-order condition (Necessary condition):

$\\\frac{dy}{dx}= 0\ or\3x^{2}-6x=0\\$

Applying the quadratic formula, we have

$\\x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\\\ 3x^{2}-6x=0\\$

Here, a= 3, b= - 6, and c= 0

$\\x=\frac{-(-6)\pm \sqrt{(-6)^{2}-4.3.0}}{2.3}\\ \\x=\frac{6\pm \sqrt{36-0}}{6}\\ \\x=\frac{6\pm \sqrt{36}}{6}\\ \\x=\frac{6\pm 6}{6}\\ \\x=\frac{6+6}{6},\frac{6-6}{6}\\ \\x=\frac{12}{6},\frac{0}{6}\\ \\x=2,0\\$

∴ Stationary points are at x= 2 and x= 0.

Second derivative:

$\\\\\frac{d^{2}y}{dx^{2}}=\frac{d}{dx} (3x^{2}-6x)\\ \\\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}3x^{2}-\frac{d}{dx}6x\\ \\\frac{d^{2}y}{dx^{2}}=3\frac{d}{dx}x^{2}-6\frac{d}{dx}x\\ \\\frac{d^{2}y}{dx^{2}}=3.2x^{2-1}-6.1\\ \\\frac{d^{2}y}{dx^{2}}=6x-6\\$