Apply maxima and minima of functions. Today I’ll show how to apply maxima and minima of functions. Have a look!!

If you’re looking for more in maxima and minima of functions, do check out:

* How to get the maximum and minimum values of functions*.

**Apply maxima and minima of functions**

**Examples of how to apply maxima and minima of functions**

Note: None of these examples are mine. I have chosen these from some book or books. I have also given the due reference at the end of the post.

So here is the first example.

**Example 1**

According to Stroud and Booth (2013)*, “The signalling range of a submarine cable is proportional to , where is the ratio of the radii of the conductor and cable. Find the value of for maximum range.”

**Solution**

Now here the signalling range of a submarine cable is proportional to . Also, is the ratio of the radii of the conductor and cable.

So that means

(1)

where is a constant greater than . Now I have to find out the value of for maximum range, that is, for the maximum value of .

**Step 1**

First of all, I’ll differentiate equation (1) with respect to . And for that I’ll use the * product rule of differentiation*. So that gives

Next, I’ll simplify it to get

(2)

Now I’ll differentiate equation (2) with respect to . Again I’ll use the same product rule of differentiation. So I get

Then I’ll simplify it to get

(3)

Now for any maximum or minimum value of , will be .

**Step 2**

Thus I can say, I’ll equate equation (2) with . So this gives

And that means either or, . But for . So, I can say

Now I’ll substitute in equation (3) to get

Since is positive, is always negative. Therefore for , is always negative and thus is maximum.

Next, I’ll get the value of for which is maximum. That is . Now I can also rewrite it as

Then I’ll take anti-log on both sides to get

So this means

Hence I can conclude that for the range is maximum. And this is the answer to the given example.

Now I’ll give another example.

**Example 2**

According to Stroud and Booth (2013)*, “The power transmitted by a belt drive is proportional to , where speed of the belt, tension on the driving side and weight per unit length of the belt. Find the speed at which the transmitted power is a maximum.”

**Solution**

Now in this example, the power transmitted by a belt drive is proportional to , where speed of the belt, tension on the driving side and weight per unit length of the belt.

So, let’s say as the power transmitted. Therefore I can say

(4)

Now I have to find out the value of the speed for which the power is maximum.

**Step 1**

First of all, I’ll differentiate equation (4) with respect to . So that gives

(5)

Next, I’ll differentiate equation (5) with respect to . So that gives

(6)

Now for any maximum or minimum value of , will be .

**Step 2**

Thus I can say, I’ll equate equation (5) with . So I get

And that gives

So this means

Next, I’ll substitute in equation (6) to get

So I can say

And this gives

Since is always positive, the value of is always negative and thus is maximum.

Hence I can conclude that for , the power transmitted is maximum. And this is the answer to the given example.

Now I’ll give another example.

**Example 3**

According to Stroud and Booth (2013)*, “The motion of a particle performing damped vibrations is given by being the displacement from its mean position at time . Show that is a maximum when and determine this maximum displacement to three significant figures.”

**Solution**

Now in this example, the motion of a particle performing damped vibrations is

(7)

Also, is the displacement of the particle from its mean position at time . And I have to prove that for , is maximum. Also, I have to get the maximum value of . So I’ll start with the differentiation.

**Step 1 **

First of all, I’ll differentiate equation (7) with respect to . And that gives

(8)

Then I’ll differentiate equation (8) with respect to to get

(9)

Now for any maximum or minimum value of , will be .

**Step 2**

Thus I can say, I’ll equate equation (8) with . So I get

And that gives either or . But for , . So this won’t work. Now for , .

Next, I’ll replace with in equation (9) to get

Then I’ll simplify it to get

Thus I can say that the value of is maximum when .

Now means

So I can say

Hence I have proved that for , the displacement is maximum.

Next, I’ll get the maximum value of for .

**Step 3 **

So means

Next, I’ll substitute this value of in equation (7) to get the value of as

Thus the maximum displacement for is upto three significant figures.

Hence I can conclude that this is the answer to the given example.

Dear friends, this is the end of today’s post. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

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