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4) A company manufactures widgets. The expense function E=8q+12,000 and the
demand function q= -10p +20,000, where p is the price.

a. Express the variable expense V in terms of q, where q is the quantity of widgets
produced

b. Express the total expenso E in terms of p.

c. Determine the Revenue R in terms of p. (Hint: R=p•q).

d. Determine the profit P in terms of p. (Hint: P=R-E).

e. Determine the price that will yield the maximum profit. (Hint: x = -b/2a).

f. Determine the maximum profit.

Respuesta :

MrRoyal

Revenue, profit and expenses are all integral parts of a company's balance sheet

  • The expense in terms of q is [tex]\mathbf{E = 8q + 12000}[/tex]
  • The expense in terms of p is [tex]\mathbf{E = -80p + 172000}[/tex]
  • The revenue in terms of p is [tex]\mathbf{R = -10p^2 + 20000p}[/tex]
  • The profit in terms of p [tex]\mathbf{P =-10p^2 + 20080p - 172000}[/tex]
  • The price that yields the maximum profit is $1004
  • The maximum profit is $9908160

The expense function is given as:

[tex]\mathbf{E = 8q + 12000}[/tex]

The demand function is given as:

[tex]\mathbf{p = -10p + 20000}[/tex]

(a) The expense in terms of q

This is already given as: [tex]\mathbf{E = 8q + 12000}[/tex]

(b) The expense in terms of p

Substitute -10p + 20000 for p in [tex]\mathbf{E = 8q + 12000}[/tex]

[tex]\mathbf{E = 8(-10p + 20000) + 12000}[/tex]

Expand

[tex]\mathbf{E = -80p + 160000 + 12000}[/tex]

[tex]\mathbf{E = -80p + 172000}[/tex]

(c) The revenue in terms of P

This is calculated as:

[tex]\mathbf{R = p \times q}[/tex]

So, we have:

[tex]\mathbf{R = p \times (-10p + 20000)}[/tex]

Expand

[tex]\mathbf{R = -10p^2 + 20000p}[/tex]

(d) The profit in terms of p

We have:

[tex]\mathbf{R = -10p^2 + 20000p}[/tex] and [tex]\mathbf{E = -80p + 172000}[/tex]

So, the profit P is calculated as:

[tex]\mathbf{P =R - E}[/tex]

This gives

[tex]\mathbf{P =-10p^2 + 20000p +80p - 172000}[/tex]

[tex]\mathbf{P =-10p^2 + 20080p - 172000}[/tex]

(e) The price that yields maximum profit

In (d), we have:

[tex]\mathbf{P =-10p^2 + 20080p - 172000}[/tex]

The maximum price is calculated using:

[tex]\mathbf{p = -\frac{b}{2a}}[/tex]

Where:

[tex]\mathbf{b =20080}[/tex]

[tex]\mathbf{a = -10}[/tex]

So, the equation becomes

[tex]\mathbf{p = -\frac{20080}{2 \times -10}}[/tex]

[tex]\mathbf{p = 1004}[/tex]

Hence, the price that yields the maximum profit is $1004

(f) The maximum profit

In (e), we have: [tex]\mathbf{p = 1004}[/tex]

Substitute 1004 for p in [tex]\mathbf{P =-10p^2 + 20080p - 172000}[/tex]

[tex]\mathbf{P =-10(1004)^2 + 20080(1004) - 172000}[/tex]

[tex]\mathbf{P =9908160}[/tex]

Hence, the maximum profit is $9908160

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