a The weekly revenue

R

of the company is given by this function, where

p

is the price of the company's product.

R = - p^{2} + 30 p

If the company hires a a distributor, they will have to pay them

4 p + 25

per week. If we let

C

equal this cost, we can model our situation using a system of equations.

{R = - p^{2} + 30 p C = 4 p + 25 (I) (II)

To find the profit of the company after hiring the distributor, we can subtract the cost

C

from the revenue

R .

This is the same as subtracting Equation (II) from Equation (I).

{R = - p^{2} + 30 p C = 4 p + 25 (I) (II)

SysEqnSub

$(I):$ Subtract $II$

{R - C = - p^{2} + 30 p - 4 p - 25 C = 4 p + 25

SubTerm

Subtract term

{R - C = - p^{2} + 26 p - 25 C = 4 p + 25

We want to find the values of

p

for which the profit,

R - C,

is positive — this means the product is profitable.

- p^{2} + 26 p - 25 > 0

Before solving the inequality, let's factor the quadratic function.

- p^{2} + 26 p - 25 = 0

UseQuadForm

Use the Quadratic Formula: $a = - 1, b = 26, c = - 25$

p_{1, 2} = \frac{- 2 6 \pm 2 6 ^{2} - 4 ( - 1 ) ( - 2 5 )}{2 ( - 1 )}

▼

Solve using the quadratic formula

CalcPowProd

Calculate power and product

p_{1, 2} = \frac{- 2 6 \pm 6 7 6 - 1 0 0}{- 2}

SubTerm

Subtract term

p_{1, 2} = \frac{- 2 6 \pm 5 7 6}{- 2}

CalcRoot

Calculate root

p_{1, 2} = \frac{- 2 6 \pm 2 4}{- 2}

ReduceFrac

$\frac{a}{b} = \frac{a / ( - 2 )}{b / ( - 2 )}$

p_{1, 2} = 13 \pm 12

The solutions are

p_{1} = 1

and

p_{2} = 25 .

Therefore, we can rewrite the profit function in factored form.

- (p - 1) (p - 25)

The graph of this function is a parabola that opens downward and intercepts the

p -

axis at

1

and

25 .

To find the values for which $- (p - 1) (p - 25) > 0,$ look at the parts of the graph that are above the $p -$ axis. From the graph we can conclude that the revenue is positive for $1 < p < 25 .$ Consequently, the product can have a price between $$ 1$ and $$ 25$ and still be profitable for the company.

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