Chapter 9.5, Problem 24E

a.

To determine

To determine when were about 15 tons of trout .

Answer to Problem 24E

15 tons of trout will be in year 1999 and 2011.

Explanation of Solution

Given:

Equation:

  $y=-0.08x^2+1.6x+10$

Calculation:

According to question:

  $y=15$

So,

  $15=-0.08x^2+1.6x+10$

Converting the equation in standard form:

  $0.08x^2-1.6x-10+15=0$

So,

  $0.08x^2-1.6x+5=0$

On comparing the equation with $a{x}^2+bx+c$

  $a=0.08,b=-1.6,c=5$

Finding the value of $x$

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substituting the values,

  $x=\frac{-(-1.6)\pm \sqrt{{(-1.6)}^2-4(0.08)(5)}}{2(0.08)}$

Opening brackets,

  $x=\frac{1.6\pm \sqrt{2.56-1.6}}{0.16}$

So,

  $x=\frac{1.6\pm \sqrt{0.96}}{0.16}$

Solving root,

  $x=\frac{1.6\pm 0.97}{0.16}$

Therefore,

  $x=\frac{1.6+0.97}{0.16}$ or $x=\frac{1.6-0.97}{0.16}$

So,

  $x=\frac{2.57}{0.16}$ or $x=\frac{0.63}{0.16}$

Therefore,

  $x=16.06$ or $x=3.93$

Therefore 15 tons of trout will be in year,

1995+3.93=1998.93 $\simeq$ 1999

And

1995+16.06=2011.06 $\simeq$ 2011

b.

To determine

To determine if this model can be used in future years.

Answer to Problem 24E

No, this model cannot be used in future years.

Explanation of Solution

Given:

Equation:

  $y=-0.08x^2+1.6x+10$

Calculation:

No, this model cannot be used in future years because this is modeled only for years between 1995-2014.