Question

If the total cost function for a commodity is given by C = 0.25x2 + 4x + 100 dollars, where x represents the number of units produced, producing how many units will result in a minimum average cost per unit? Find the minimum average cost

Answer 1

$\large\underline{\sf{Solution-}}$

Given Cost Function for a commodity is

$\rm :\longmapsto\:C = 0.25 {x}^{2} + 4x + 100$

So, Average Cost of the commodity is given by

$\rm :\longmapsto\:AC = \dfrac{C}{x}$

$\rm :\longmapsto\:AC = \dfrac{0.25 {x}^{2} + 4x + 100 }{x}$

$\rm :\longmapsto\:AC = 0.25x + 4 + \dfrac{100}{x}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}AC =\dfrac{d}{dx} \bigg(0.25x + 4 + \dfrac{100}{x}\bigg)$

$\rm :\longmapsto\:\dfrac{d}{dx}AC =0.25\dfrac{d}{dx}x + \dfrac{d}{dx}4 + 100\dfrac{d}{dx} {x}^{ - 1}$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}k = 0}}}$

and

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} }}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{d}{dx}AC =0.25 + 0 + 100( - 1) {x}^{ - 1 - 1}$

$\rm :\longmapsto\:\dfrac{d}{dx}AC =0.25 - 100 {x}^{ - 2}$

For maxima or minima,

$\rm :\longmapsto\:\dfrac{d}{dx}AC =0$

$\rm :\longmapsto\:0.25 - 100 {x}^{ - 2} = 0$

$\rm :\longmapsto\:\dfrac{25}{100} = \dfrac{100}{ {x}^{2} }$

$\rm :\longmapsto\:\dfrac{1}{4} = \dfrac{100}{ {x}^{2} }$

$\rm :\longmapsto\: {x}^{2} = 400$

$\bf\implies \:x = 20 \: units$

As we have,

$\rm :\longmapsto\:\dfrac{d}{dx}AC =0.25 - 100 {x}^{ - 2}$

So, on differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{ {d}^{2} }{d {x}^{2} }AC =\dfrac{d}{dx}(0.25 - 100 {x}^{ - 2} )$

$\rm :\longmapsto\:\dfrac{ {d}^{2} }{d {x}^{2} }AC =\dfrac{d}{dx}0.25 - \dfrac{d}{dx}100 {x}^{ - 2}$

$\rm :\longmapsto\:\dfrac{ {d}^{2} }{d {x}^{2} }AC =0 - 100\dfrac{d}{dx}{x}^{ - 2}$

$\rm :\longmapsto\:\dfrac{ {d}^{2} }{d {x}^{2} }AC =- 100( - 2) {x}^{ - 2 - 1}$

$\rm :\longmapsto\:\dfrac{ {d}^{2} }{d {x}^{2} }AC =200 {x}^{ -3} > 0$

$\bf\implies \:AC \: is \: minimum \: at \: x = 20$

And Minimum Average Cost is

$\rm :\longmapsto\:AC _{{x = 20}} = 0.25 \times 20 + 4 + \dfrac{100}{20}$

$\rm :\longmapsto\:AC _{{x = 20}} = 5 + 4 + 5$

$\rm :\longmapsto\:AC _{{x = 20}} = 14$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Basic Concept Used :-

Let y = f(x) be a given function.

To find the maximum and minimum value, the following steps are follow :

1. Differentiate the given function.

2. For maxima or minima, put f'(x) = 0 and find critical points.

3. Then find the second derivative, i.e. f''(x).

4. Apply the critical points ( evaluated in second step ) in the second derivative.

5. Condition :-

• The function f (x) is maximum when f''(x) < 0.

• The function f (x) is minimum when f''(x) > 0.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

LEARN MORE

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$