Question

The annual inventory cost C for a manufacturer is given below, where Q is the order size when the inventory is replenished. Find the change in annual cost when Q is increased from 347 to 348, and compare this with the instantaneous rate of change when Q = 347. (Round your answers to two decimal places.) C = 1,017,000/ Q + 6.8QChange in C= ___________C(360)= ___________

Answer 1

Answer:

Change in annual cost is 1.63 (decreasing).

Instantaneous rate of change is 1.65 (decreasing).

Step-by-step explanation:

Given that,

$C=\dfrac{1017000}{Q}+6.8\:Q$

The annual change in cost is given by,  

$Change\:in\:C=C\left(348\right)-C\left(347\right)$

Calculate the value of cost at Q = 347 and Q =348 by substituting the value in cost function.

Calculating value of C at Q=347,

$C=\dfrac{1017000}{347}+6.8\left(347\right)$

$C=5290.44$

Calculating value of C at Q=348,

$C=\dfrac{1017000}{348}+6.8\left(348\right)}$

$C= 5288.81$

Substituting the value,  

$Change\:in\:C=5288.81-5290.44$

$Change\:in\:C=-1.63$

Negative sign indicate that there is decrease in annual cost when Q is increased from 347 to 348

Therefore, change is annual cost is 1.63  (decreasing).

Instantaneous rate of change is given by the formula,  

$f’\left(x\right)=\lim_{h\rightarrow 0}\dfrac{f\left(x+h\right)-f\left(x\right)}{h}$

Rewriting,

$C’\left(Q\right)=\lim_{h\rightarrow 0}\dfrac{C\left(Q+h\right)-C\left(Q\right)}{h}$

Now calculate \dfrac{C\left(Q+h\right) by substituting the value Q+h in cost function,

$C\left(Q+h\right)= \dfrac{1017000}{Q+h}+6.8\left(Q+h\right)$

Therefore,  

$C’\left(Q\right)= \lim _{h\to 0}\:\dfrac{\dfrac{1017000}{Q+h}+6.8\left(Q+h\right)-\left(\dfrac{1017000}{Q}+6.8Q\right)}{h}$

By using distributive law,

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000}{Q+h}+6.8Q+6.8h-\dfrac{1017000}{Q}-6.8Q}{h}$

Cancelling out common factors,  

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000}{Q+h}-\dfrac{1017000}{Q}+6.8h}{h}$

Now, LCD of $\left(Q+h\right)$ and $Q$ is $Q(Q+h)$. So multiplying first term by $Q$  and second term by

Therefore,  

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000}{Q+h}\times\dfrac{Q}{Q}-\dfrac{1017000}{Q}\times\dfrac{Q+h}{Q+h}+6.8h}{h}$

Simplifying,  

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000Q}{\left(Q+h\right)\left(Q\right)}-\dfrac{1017000\left(Q+h\right)}{Q\left(Q+h\right)}+6.8h}{h}$

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000Q-1017000\left(Q+h\right)}{\left(Q+h\right)\left(Q\right)}+6.8h}{h}$

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{1017000Q-1017000Q-1017000h}{\left(Q+h\right)\left(Q\right)}+6.8h}{h}$

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{\dfrac{-1017000h}{\left(Q+h\right)\left(Q\right)}+6.8h}{h}$

Factoring out h from numerator,  

$C’\left(Q\right) = \lim _{h\to 0}\:\dfrac{h\left(\dfrac{-1017000}{\left(Q+h\right)\left(Q\right)}+6.8\right)}{h}$

Cancelling out h,  

$C'\left(Q\right) = \lim_{h\to 0}\: -\dfrac{1017000}{\left(Q+h\right)Q}+6.8$

Calculating the limit by plugging value h = 0,

$C'\left(Q\right) = -\dfrac{1017000}{\left(Q+0\right)Q}+6.8$

$C'\left(Q\right) = -\dfrac{1017000}{Q^{2}}+6.8$

Given that Q=347,

$C'\left(347\right) = -\dfrac{1017000}{347^{2}}+6.8$

$C'\left(347\right) = -1.65$

Negative sign indicate that there is decrease in instantaneous rate when Q is 347

Therefore, instantaneous rate of change at Q=347 is 1.65  (decreasing).