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 August 2nd, 2011, 04:41 PM #1 Newbie   Joined: Aug 2011 Posts: 12 Thanks: 0 I need help with this The profit function is given by P= -2X(square) + 72X-198 Where X units is the quantity and P the profit. Find the maximum profit and the number of units where the maximum profit occurs. Sketch the curve of the profit function.
 August 2nd, 2011, 05:57 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: I need help with this We are given: $P(x)=-2x^2+72x-198$ This is a downward opening parabola, and to find the point of maximum profit, we need to find the vertex of the parabola (h,k) by writing it in the form: $P(x)=a(x-h)^2+k$ Recall, the vertex of a parabola is where its global max/min occurs. Equating the two expressions, we find: $-2x^2+72x-198=a(x-h)^2+k=a$$x^2-2hx+h^2$$+k=ax^2-2ahx+ah^2+k$ Equating coefficients, we find: $a=-2$ $-2ah=4h=72\:\therefore\:h=18$ $ah^2+k=-648+k=-198\:\therefore\:k=450$ thus, we may write: $P(x)=-2(x-1^2+450" /> Thus, maximum profit is 450 at a quantity of 18. To sketch the curve, begin by plotting the point (18,450). We may get 2 more points by finding the roots, or x-intercepts. We do this by setting P(x) = 0. $-2(x-1^2+450=0" /> $(x-1^2-225=0" /> $(x-1^2=225" /> $x-18=\pm15$ $x=18\pm15$ Thus, one root is $18-15=3$ and the other is $18+15=33$. So now plot the points (3,0) and (33,0) and connect the 3 points with a parabolic curve. Notes: We may also have found the vertex by completing the square: $P(x)=-2x^2+72x-198=-2(x-36)-198=-2$$x-36+18^2$$-198+2$$18^2$$=-2$$x-18$$^2+450$ Or, we could use differential calculus: $P(x)=-2x^2+72x-198$ $P'(x)=-4x+72=0\:\therefore\=18" /> $P(1=-2(1^2+72(1-198=450" /> To find the roots, we could have factored: $P(x)=-2x^2+72x-198=-2$$x^2-36x+99$$=-2(x-3)(x-33)$ thus we see the roots are x = 3, 33. Or we could have used the quadratic formula on $x^2-36x+99=0$: $x=\frac{36\pm\sqrt{36^2-4\cdot99}}{2}=\frac{36\pm6\sqrt{36-11}}{2}=18\pm3\sqrt{25}=18\pm15$
 August 2nd, 2011, 07:06 PM #3 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: I need help with this Or... $\left( \frac{-b}{2a}, P(\frac{-b}{2a}) \right )$
 August 2nd, 2011, 07:11 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: I need help with this You know, I actually thought about adding the axis of symmetry, but didn't.
August 2nd, 2011, 07:29 PM   #5
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Re: I need help with this

Quote:
 Originally Posted by MarkFL You know, I actually thought about adding the axis of symmetry, but didn't.
I myself have been known to dabble in minimalism from time to time.

 August 2nd, 2011, 07:37 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: I need help with this Although...you know...it was kind of implied in the differential calculus method, but one need only look at the quadratic formula to easily see where the axis of symmetry lies. There's nothing wrong with minimalism. :P

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