the profit function for a bussiness is given by the equation p(x)=-4x^2+16x-7,where x is the number of items sold, in thousands, and p(x) is dollars in thousands. calculate the maximum profit and how many items must be sold to acieve it?

bioplant
Oct 14, 2018

#1**+2 **

The maximum or minimum of any quadratic is always at the input value of \(\frac{-b}{2a}\) where the standard form of a quadratic is \(ax^2+bx+c=0\) .

\(a=-4, b=16;\\ x=\frac{-b}{2a}\) | Substitute in the known values for a and b . |

\(x=\frac{-16}{2*-4}\) | Solve for x by simplifying completely. |

\(x=2\) | This is the optimal number of items sold, in thousands, in order to make maximum profit. Let's substitute that into the original function to determine the profit with that number of items sold. |

\(x=2;\\ p(2)=-4*2^2+16*2-7\) | Yet again, just simplify from here. |

\(p(2)=-4*4+32-7\) | |

\(p(2)=-16+25\) | |

\(p(2)=9\) | This is the maximum profit that this business, in thousands of dollars, makes. |

TheXSquaredFactor
Oct 14, 2018

#1**+2 **

Best Answer

The maximum or minimum of any quadratic is always at the input value of \(\frac{-b}{2a}\) where the standard form of a quadratic is \(ax^2+bx+c=0\) .

\(a=-4, b=16;\\ x=\frac{-b}{2a}\) | Substitute in the known values for a and b . |

\(x=\frac{-16}{2*-4}\) | Solve for x by simplifying completely. |

\(x=2\) | This is the optimal number of items sold, in thousands, in order to make maximum profit. Let's substitute that into the original function to determine the profit with that number of items sold. |

\(x=2;\\ p(2)=-4*2^2+16*2-7\) | Yet again, just simplify from here. |

\(p(2)=-4*4+32-7\) | |

\(p(2)=-16+25\) | |

\(p(2)=9\) | This is the maximum profit that this business, in thousands of dollars, makes. |

TheXSquaredFactor
Oct 14, 2018