Question

Q16) The total cost function C = x3 + 3x + 4, then the marginal cost when x = 10 units is
a) 303
b) 200
c) 340
d) 250
017) TL
intorart at 8% on Rs 100 for 5 years is Rs​

Answer 1

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

Given Cost Function is

$\rm :\longmapsto\:C = {x}^{3} + 3x + 4$

We know,

Marginal Cost is defined as Rate of change of Cost Function per unit.

Mathematically, Marginal Cost ( M.C. ) is represented as

$\rm :\longmapsto\:\boxed{\tt{M.C. \: = \: \dfrac{d}{dx}C }}$

$\rm :\longmapsto\:C = {x}^{3} + 3x + 4$

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

$\rm :\longmapsto\:\dfrac{d}{dx} C = \dfrac{d}{dx}\bigg[{x}^{3} + 3x + 4\bigg]$

$\rm :\longmapsto\:M.C. = \dfrac{d}{dx}{x}^{3} +\dfrac{d}{dx} 3x + \dfrac{d}{dx}4$

We know,

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

and

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

$\rm :\longmapsto\:M.C. = {3x}^{3 - 1} + 3 \times 1 + 0$

$\rm :\longmapsto\:M.C. = {3x}^{2} + 3$

So, Marginal Cost when x = 10, we get

$\rm :\longmapsto\:M.C._{{x = 10}} = {3(10)}^{2} + 3$

$\rm :\longmapsto\:M.C._{{x = 10}} = 300 + 3$

$\rm :\longmapsto\:M.C._{{x = 10}} = 303$

• So, Option (a) is correct.

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More to Know

Average Cost

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: Average \: Cost, \: AC = \frac{Cost \: function \: }{x} \: }}}$

Revenue Function

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: Revenue, \: R \: = \:p \: x \: }}}$

Average Revenue

$\purple{\rm :\longmapsto\:\boxed{\tt{ \: \: Average \: Revenue, \: AC = \frac{Revenue \: }{x} \: }}}$

Marginal Revenue

$\blue{\rm :\longmapsto\:\boxed{\tt{M.R. \: = \: \dfrac{d}{dx}R }}}$

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