class 12 maths application of derivatives

Find the maximum profit that a company can make , if the profit function is given by $p\left( x \right) = 41 - 72x - 18{x^2}$ .

VAVidaara Admin Asked 8d ago 0 views 0 answers
📘 Application of Derivatives NCERT Ex. 6.5, Q.6,Page 232 SA

Find the maximum profit that a company can make , if the profit function is given by $p\left( x \right) = 41 - 72x - 18{x^2}$ .

Official Solution

VVidaara Team ✓ Verified solution NCERT & Exemplar

We have, $p\left( x \right) = 41 - 72x - 18{x^2}$
$\Rightarrow p'\left( x \right) = - 72 - 36x$

Now, for critical points, $p'\left( x \right) = 0$
$\Rightarrow - 72 - 36x = 0 \Rightarrow x = - 2$
$p''\left( x \right) = - 36 < 0 \Rightarrow p''\left( { - 2} \right) = - 36 < 0$

Therefore the profit is maximum at $x = - 2$ and maximum profit is
$p\left( { - 2} \right) = 41 - 72\left( { - 2} \right) - 18{\left( { - 2} \right)^2} = 41 + 144 - 72 = 185 - 72 = 113$ units

View the full step-by-step solution page & related questions →

Community Answers (0)

Log in to post your own answer or join the discussion.

Discussion (0)

No comments yet — start the discussion.

← Back to all questions