Chapter ISG, Problem 9.5.5P

a.

To determine

To calculate zeros of equation.

Answer to Problem 9.5.5P

7 seconds

Explanation of Solution

Given information:

Number of sales of company

  $f(t)=-5000x^2+70000x+5000$

Formula used:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

x are roots of quadratic equation

a, b and c are constants

Calculation:

  $f(t)=-5000x^2+70000x+5000$

For zeros

  $f(t)=0$

  $0=-5000x^2+70000x+5000$

On solving,

  $x^2-14x-1=0$

On solving,

  $x=\frac{-(-14)\pm \sqrt{{(-14)}^2-4(1)(-1)}}{2(1)}$

On solving,

  $x=\frac{14\pm \sqrt{196+4}}{2}$

On solving,

  $x=\frac{14\pm \sqrt{200}}{2}$

On solving,

  $x=\frac{14\pm 10\sqrt{2}}{2}$

On solving,

  $x=\frac{7+5\sqrt{2}}{2}$ and $x=\frac{7-5\sqrt{2}}{2}$

Approximate values $x=0$ and $x=14$

b.

To determine

To calculate maximum number of sale of company.

Answer to Problem 9.5.5P

maximum number of sale are 5000

Explanation of Solution

Given information:

Number of sales of company

  $f(t)=-5000x^2+70000x+5000$

  $x=0$ and $x=14$

Calculation:

Number of sales of company

  $f(t)=-5000x^2+70000x+5000$

As Substituting $x=14$to calculate maximum number of sales

  $f(14)=-5000{(14)}^2+70000(14)+5000$

On solving,

  $f(14)=-5000{(14)}^2+70000(14)+5000$

On solving,

  $f(14)=-5000\times 196+980000+5000$

On solving,

  $f(14)=-980000+980000+5000$

On solving,

  $f(14)=5000$

Hence, maximum number of sale are 5000