Mobile Tech manufactures cell phone accessories. A particular item in their product line costs $40 each
to manufacture. The fixed costs are $120,000. The demand function is q=-120p+8,000, where q is the
quantity the public will buy given the price p.
a.
Write the expense function in terms of p.
B. what is the revenue equation for this mobile tech product?write the revenue equation in terms of the price.
c.use the revenue equation from part d.what would the revenue be if the price of the item was set at $60?
d. determine the roots of the revenue equation.interpret those roots in the context of this problem
e. which price would yield the higher revenue, $20 or $40?

The expense function includes both the variable cost (cost to manufacture each item) and the fixed cost. The variable cost is the product of the cost to manufacture each item ($40) and the quantity ([tex]q[/tex]).

[tex] E(p) = 40q + 120,000 [/tex]

Now, we need to substitute the demand function [tex]q = -120p + 8,000[/tex] into the expense function:

[tex] E(p) = 40(-120p + 8,000) + 120,000 [/tex]

[tex] E(p) = -4,800p + 320,000 + 120,000 [/tex]

[tex] E(p) = -4,800p + 440,000 [/tex]

So, the expense function in terms of [tex]p[/tex] is:

[tex]\Large\boxed{\boxed{E(p) = -4,800p + 440,000}}[/tex]

[tex]\dotfill[/tex]

b. Revenue Equation (R):

The revenue is the product of the quantity sold ([tex]q[/tex]) and the price ([tex]p[/tex]).

[tex] R(p) = pq [/tex]

Substitute the demand function into the revenue equation:

[tex] R(p) = p(-120p + 8,000) [/tex]

[tex] R(p) = -120p^2 + 8,000p [/tex]

So, the revenue equation for this mobile tech product is:

[tex] \Large\boxed{\boxed{R(p) = -120p^2 + 8,000p}} [/tex]

[tex]\dotfill[/tex]

c. Revenue at $60 price:

Substitute [tex]p = 60[/tex] into the revenue equation:

[tex] R(60) = -120(60)^2 + 8,000(60) [/tex]

Calculate [tex]R(60)[/tex] to find the revenue at a price of $60.

Let's calculate [tex] R(60) [/tex]:

[tex]\begin{aligned} R(60) & = -120(60)^2 + 8,000(60) \\\\ & = -120(3600) + 480,000 \\\\ & = -432,000 + 480,000 \\\\ & = 48,000\\\\ & = 48,000 \end{aligned}[/tex].

The revenue if the price of the item is set at $60 is:
[tex]\Large\boxed{\boxed{\$ 48,000}}[/tex]

[tex]\dotfill[/tex]

d. Roots of the Revenue Equation:

The revenue equation is a quadratic equation, and its roots can be found by setting [tex]R(p) = 0[/tex]:

[tex] -120p^2 + 8,000p = 0 [/tex]

Factor out [tex]p[/tex]:

[tex] -120p( p - \dfrac{8,000}{120}) = 0 [/tex]

This equation has two solutions,

Either

[tex] -120 p = 0 [/tex]

[tex] p =\dfrac{0}{-120} [/tex]

[tex] p = 0[/tex]

Or

[tex] p - \dfrac{8,000}{120} = 0[/tex]

[tex] p = \dfrac{8,000}{120} [/tex]

[tex] p \approx 66.66 [/tex]

This equation has roots at:
[tex]\large\boxed{\boxed{ \sf p = 0 \textsf{ and } p =66.66 }}[/tex]

[tex]\dotfill[/tex]

e. Higher Revenue at $20 or $40:

let's compare the revenue at [tex]p = 20[/tex] and [tex]p = 40[/tex]:

[tex]\begin{aligned}R(20) &= -120(20)^2 + 8,000(20) \\\\&= -120(400) + 160,000 \\ \\ &= -48,000 + 160,000 \\\\ &= 112,000 \end{aligned}[/tex]

[tex]\begin{aligned} R(40) &= -120(40)^2 + 8,000(40) \\\\ &= -120(1600) + 320,000 \\\\ &= -192,000 + 320,000 \\\\ &= 128,000 \end{aligned}[/tex]

Therefore, the price of $40 yields higher revenue compared to $20.