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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the maximum profit that a company can make, if the profit function is given by $ p(x) = 41 - 72x - 18x^2$

$\begin{array}{1 1} 59\\ 49 \\ 60 \\100 \end{array} $

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1 Answer

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Toolbox:
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
$p(x)=-18x^2-24x+41$
$p'(x)=-36x-24$
$\qquad=-12(2+3x)$
For maxima and minima
$p'(x)=0$
Now $p'(x)=0$
$\Rightarrow 12(2+3x)=0$
$2+3x=0$
$3x=-2$
$x=\large\frac{-2}{3}$
Step 2:
$p'(x)$ changes sign from +ve to -ve.
$\Rightarrow p(x)$ has a maximum value at $x=\large\frac{-2}{3}$
Maximum profit=$p(\large\frac{-2}{3}\big)$
$\qquad\qquad\quad\;\;=41-24\big(\large\frac{-2}{3}\big)$$-18\big(\large\frac{-2}{3}\big)^2$
$\qquad\qquad\quad\;\;=41+16-8$
$\qquad\qquad\quad\;\;=49$
answered Aug 7, 2013 by sreemathi.v
 

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