N.C.E.R.T problem

Calculus Level 2

In a city the population is given by f ( x ) = 1000 + 1000 x 100 + x 2 . f(x) = 1000 + \frac{1000x}{100 + x^{2}}.

What is the maximum population of the city?


The answer is 1050.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

U Z
Nov 11, 2014

f ( x ) = 1000 + 1000 100 x + x f(x) = 1000 + \dfrac{1000}{\frac{100}{x} + x}

Population maximum when denominator minimum

100 x + x 2 100 \frac{\frac{100}{x} + x}{2} \geq \sqrt{100}

Denominator = 20

f ( x ) m a x = 1000 + 1000 20 = 1050 f(x)_{max} = 1000 + \dfrac{1000}{20} = 1050

Question asked is second derivative test?!

Murugesh M - 6 years, 7 months ago

Log in to reply

You can apply second derivative test here , as specified it was a N.C.E.R.T problem. I am sharing just my approach, just in 2 steps we get the answer

U Z - 6 years, 7 months ago

but how to configure the 1st step you stated?

Gandhar Joshi - 6 years, 7 months ago

Log in to reply

Multiply and divide by x

Vishnu Bhagyanath - 5 years, 11 months ago

Log in to reply

No no.. I mean the next one ?

Gandhar Joshi - 5 years, 11 months ago
Gandhar Joshi
Nov 14, 2014

just the normal maxima test.... careful! the equations are too dizzy!!!

Can you show the steps?

Calvin Lin Staff - 6 years, 6 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...