Economics at Zero Marginal Cost: Maxima and Minima of a Quartic Function (Polynomial Function of degree 4)

Find the maximum and minimum value of the function y=3x⁴−10x³+8x²+4

Solution:

Given: y=3x⁴−10x³+8x²+4

To find: The maximum and minimum value of the function

First derivative:

$\\\\\frac{dy}{dx}=\frac{d}{dx} (3x^{4}-10x^{3}+8x^{2}+4)\\ \frac{dy}{dx}=\frac{d}{dx}3x^{4}-\frac{d}{dx} 10x^{3}+\frac{d}{dx}8x^{2}+\frac{d}{dx}4\\ \frac{dy}{dx}=3\frac{d}{dx}x^{4}-10\frac{d}{dx}x^{3}+8\frac{d}{dx}x^{2}+0\\ \frac{dy}{dx}=3.4x^{4-1}-10.3x^{3-1}+8.2x^{2-1}\\ \frac{dy}{dx}=12x^{3}-30x^2+16x\\$

First-order condition (Necessary condition):

$\\\\\frac{dy}{dx}=0\ or\12x^{3}-30x^2+16x=0\\$

Taking x as a common factor

$\\x(12x^{2}-30x+16)=0\\ Either\ x=0\ or \12x^{2}-30x+16\\$

Applying the quadratic formula, we have

$\\x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\ 12x^{2}-30x+16=0\\$

Here, a=12, b=-30, and c=16

$\\x=\frac{-(-30))\pm \sqrt{(-30)^{2}-4.12.16}}{2.12}\\ \\x=\frac{30\pm \sqrt{900-768}}{24}\\ \\x=\frac{30\pm \sqrt{132}}{24}\\ \\x=\frac{30\pm \11.489}{24}\\ \\x=\frac{30+11.489}{24},\frac{30-11.489}{24}\\ \\x=\frac{41.489}{24},\frac{18.511}{24}\\ \\x=1.729,0.771\\$

∴ Stationary points are at x=0, x=1.729 and x=0.771.

Second derivative:

$\\\\\frac{d^{2}y}{dx^{2}}=\frac{d}{dx} (12x^{3}-30x^2+16x)\\ \\\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}12x^{3}-\frac{d}{dx}30x^2+\frac{d}{dx}16x\\ \\\frac{d^{2}y}{dx^{2}}=12\frac{d}{dx}x^{3}-30\frac{d}{dx}x^{2}+16\frac{d}{dx}x\\ \\\frac{d^{2}y}{dx^{2}}=12.3x^{3-1}-30.2x^{2-1}+16.1\\ \\\frac{d^{2}y}{dx^{2}}=36x^{2}-60x+16\\$

Second-order condition (Sufficient condition):

$\\At\ x=0\\\\ \frac{d^{2}y}{dx^{2}}=36x^{2}-60x+16=36(0)^{2}-60(0)+16=36(0)-0+16=0+16=16>0\\ \\At\ x=1.729\\\\ \frac{d^{2}y}{dx^{2}}=36x^{2}-60x+16=36(1.729)^{2}-60(1.729)+16=36(2.989)-103.74+16=107.604-103.74+16=19.864>0\\ \\At\ x=0.771\\\\ \frac{d^{2}y}{dx^{2}}=36x^{2}-60x+16=36(0.771)^{2}-60(0.771)+16=36(0.594)-46.26+16=21.384-46.26+16=-8.876<0\\$

Hence, the function y=3x⁴−10x³+8x²+4 is minimum at x=0 and x=1.729 and maximum at x=0.771.

$\\y_{min}=3x^{4}-10x^{3}+8x^{2}+4\\ y_{min}=3(0)^{4}-10(0)^{3}+8(0)^{2}+4\\ y_{min}=3(0)-10(0)+8(0)+4\\ y_{min}=0-0+0+4\\ y_{min}=4\\ \\y_{min}=3x^{4}-10x^{3}+8x^{2}+4\\ y_{min}=3(1.729)^{4}-10(1.729)^{3}+8(1.729)^{2}+4\\ y_{min}=3(8.937)-10(5.169)+8(2.989)+4\\ y_{min}=26.811-51.69+23.912+4\\ y_{min}=3.033\\ \\y_{max}=3x^{4}-10x^{3}+8x^{2}+4\\ y_{max}=3(0.771)^{4}-10(0.771)^{3}+8(0.771)^{2}+4\\ y_{max}=3(0.353)-10(0.458)+8(0.594)+4\\ y_{max}=1.059-4.58+4.752+4\\ y_{max}=5.231\\$