Exercise 49 Page 197

The graph below illustrates the profit of a company in terms of the selling price of a certain product. The blue part represents positive profit and the red part represents negative profit. We can see that there are two critical selling price values.

These critical values tell the company that if they ask too little or too much for their product, they will lose money. To answer the question we need to find the second critical value — the value when the profit is 0. - 8x^2+165x-100=0We can use the Quadratic Formula to solve this quadratic equation. 2 &Quadratic equation: && ax^2+ bx+ c=0 &Solutions:&&x=- b±sqrt(b^2-4 a c)/2 a Let's identify the coefficients in our quadratic expression. - 8x^2+165x-100= - 8x^2+ 165x+( - 100) We see that a= - 8, b= 165, and c= -100. Let's substitute these values into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 165±sqrt(165^2-4( -8)( -100))/2( -8)

▼

Simplify right-hand side

CalcPow

Calculate power

x=- 165±sqrt(27225-4( -8)( -100))/2( -8)

MultPosNeg

a(- b)=- a * b

x=- 165±sqrt(27225-(-32)( -100))/-16

MultNegNegTwoPar

(- a)(- b)=a* b

x=- 165±sqrt(27225-3200)/-16

SubTerms

Subtract terms

x=- 165±sqrt(24025)/-16

CalcRoot

Calculate root

x=-165±155/-16

We found two solutions that correspond to 0 profit. -165+155/-16&=-10/-16=0.625 -165-155/-16&=-320/-16=20 The maximum price per unit that the company can charge without losing money is $20. The correct answer is B.