Chapter 4.4, Problem 60E

(a.)

To determine

The quantity that will minimize $A\left(q\right)$ .

Answer to Problem 60E

It has been determined that the quantity that will minimize $A\left(q\right)$ is given as $q=\sqrt{\frac{2km}{h}}$ .

Explanation of Solution

Given:

The average weekly cost of ordering, paying for, and holding merchandise is,

  $A\left(q\right)=\frac{km}{q}+cm+\frac{hq}{2}$

Here, $q$ is the quantity ordered when things run low, $k$ is the cost of placing an order (constant), $c$ is the cost of one item (constant), $m$ is the number of items sold each week (constant), $h$ is the weekly holding cost per item (constant).

Concept used:

A function attains its minimum value at a point for which the first derivative of the function is zero and the second derivative of the function is positive.

Calculation:

It is given that $A\left(q\right)=\frac{km}{q}+cm+\frac{hq}{2}$ .

Differentiating with respect to $q$ ,

  $A^{\prime }\left(q\right)=-\frac{km}{q^2}+\frac{h}{2}$

Differentiating again with respect to $q$ ,

  $A^{″}\left(q\right)=\frac{km}{q^3}$

Equating first derivative to zero,

  $-\frac{km}{q^2}+\frac{h}{2}=0$

Simplifying,

  $\frac{km}{q^2}=\frac{h}{2}$

On further simplification,

  $q^2=\frac{2km}{h}$

Solving,

  $q=\pm \sqrt{\frac{2km}{h}}$

Now, $q$ is the quantity ordered when things run low. So, $q$ cannot be negative.

Considering the positive root only,

  $q=\sqrt{\frac{2km}{h}}$

Put $q=\sqrt{\frac{2km}{h}}$ in $A^{″}\left(q\right)=\frac{km}{q^3}$ to get,

  $A^{″}\left(q\right)=\frac{km}{{\left(\frac{2km}{h}\right)}^{\frac{3}{2}}}>0$

Note that the above quantity is positive because $k$ is the cost of placing an order (constant), $m$ is the number of items sold each week (constant) and $h$ is the weekly holding cost per item (constant); which are all positive quantities.

So, the quantity that will minimize $A\left(q\right)$ is given as $q=\sqrt{\frac{2km}{h}}$ .

Conclusion:

It has been determined that the quantity that will minimize $A\left(q\right)$ is given as $q=\sqrt{\frac{2km}{h}}$ .

(b.)

To determine

The most economical quantity to order when $k$ is replaced by $k+bq$ , where $b$ is a constant.

Answer to Problem 60E

It has been determined that the most economical quantity to order, when $k$ is replaced by $k+bq$ , where $b$ is a constant; is still given as $q=\sqrt{\frac{2km}{h}}$ .

Explanation of Solution

Given:

The average weekly cost of ordering, paying for, and holding merchandise is,

  $A\left(q\right)=\frac{km}{q}+cm+\frac{hq}{2}$

Here, $q$ is the quantity ordered when things run low, $k$ is the cost of placing an order (constant), $c$ is the cost of one item (constant), $m$ is the number of items sold each week (constant), $h$ is the weekly holding cost per item (constant).

Concept used:

A function attains its minimum value at a point for which the first derivative of the function is zero and the second derivative of the function is positive.

Calculation:

It is given that $A\left(q\right)=\frac{km}{q}+cm+\frac{hq}{2}$ .

Replacing $k$ by $k+bq$ , where $b$ is a constant, in the above expression,

  $A_1\left(q\right)=\frac{\left(k+bq\right)m}{q}+cm+\frac{hq}{2}$

Simplifying,

  $A_1\left(q\right)=\frac{km}{q}+bm+cm+\frac{hq}{2}$

Differentiating with respect to $q$ ,

  $A_1{}^{\prime }\left(q\right)=-\frac{km}{q^2}+\frac{h}{2}$

Differentiating again with respect to $q$ ,

  $A_1{}^{\prime \text{}\prime }\left(q\right)=\frac{km}{q^3}$

Equating first derivative to zero,

  $-\frac{km}{q^2}+\frac{h}{2}=0$

Simplifying,

  $\frac{km}{q^2}=\frac{h}{2}$

On further simplification,

  $q^2=\frac{2km}{h}$

Solving,

  $q=\pm \sqrt{\frac{2km}{h}}$

Now, $q$ is the quantity ordered when things run low. So, $q$ cannot be negative.

Considering the positive root only,

  $q=\sqrt{\frac{2km}{h}}$

Put $q=\sqrt{\frac{2km}{h}}$ in $A_1{}^{\prime \text{}\prime }\left(q\right)=\frac{km}{q^3}$ to get,

  $A_1{}^{\prime \text{}\prime }\left(q\right)=\frac{km}{{\left(\frac{2km}{h}\right)}^{\frac{3}{2}}}>0$

Note that the above quantity is positive because $k$ is the cost of placing an order (constant), $m$ is the number of items sold each week (constant) and $h$ is the weekly holding cost per item (constant); which are all positive quantities.

So, the quantity that will minimize $A_1\left(q\right)$ is given as $q=\sqrt{\frac{2km}{h}}$ .

Thus, the most economical quantity to order, when $k$ is replaced by $k+bq$ , where $b$ is a constant; is still given as $q=\sqrt{\frac{2km}{h}}$ .

Conclusion:

It has been determined that the most economical quantity to order, when $k$ is replaced by $k+bq$ , where $b$ is a constant; is still given as $q=\sqrt{\frac{2km}{h}}$ .