Question

consider the cost function ( c(x) = 8x^2 + 12x + 25 ) (thousand dollars).
a) what is the marginal cost at production level ( x = 5 )?
b) use the marginal cost at ( x = 5 ) to estimate the cost of producing 5.25 units.
c) let ( r(x) = -x^2 + 61x + 55 ) denote the revenue in thousands of dollars generated from the production of ( x ) units. what is the break - even point? (recall that the break - even point is when the revenue is equal to the cost.)
d) compute and compare the marginal revenue and marginal cost at the break - even point. should the company increase production beyond the break - even point?

Explanation:

Part (a)

Step1: Recall marginal cost formula

The marginal cost \( MC(x) \) is the derivative of the cost function \( C(x) \). Given \( C(x) = 8x^2 + 12x + 25 \), we find its derivative using the power rule. The power rule states that if \( f(x) = ax^n \), then \( f'(x) = nax^{n - 1} \).

Step2: Compute the derivative

For \( C(x) = 8x^2 + 12x + 25 \), the derivative \( C'(x) \) (which is the marginal cost function \( MC(x) \)) is:
\( C'(x)=\frac{d}{dx}(8x^2)+\frac{d}{dx}(12x)+\frac{d}{dx}(25) \)
Using the power rule:

• For \( 8x^2 \), \( n = 2 \), \( a = 8 \), so the derivative is \( 2\times8x^{2 - 1}=16x \)
• For \( 12x \), \( n = 1 \), \( a = 12 \), so the derivative is \( 1\times12x^{1 - 1}=12 \)
• The derivative of a constant (25) is 0.

So, \( MC(x)=C'(x)=16x + 12 \)

Step3: Evaluate at \( x = 5 \)

Substitute \( x = 5 \) into the marginal cost function:
\( MC(5)=16\times5+12 \)
\( MC(5)=80 + 12=92 \) (in thousand dollars per unit)

Step1: Recall the approximation using marginal cost

The marginal cost at \( x = 5 \) gives the approximate change in cost when production is increased by 1 unit from \( x = 5 \) to \( x = 6 \). To estimate the cost of producing \( 5.25 \) units, we consider the change in \( x \) is \( \Delta x=5.25 - 5 = 0.25 \)

Step2: Use the marginal cost for approximation

The change in cost \( \Delta C\approx MC(5)\times\Delta x \)
We know \( MC(5) = 92 \) (from part (a)) and \( \Delta x=0.25 \)
So, \( \Delta C\approx92\times0.25 = 23 \)
First, we find the cost at \( x = 5 \): \( C(5)=8\times5^2+12\times5 + 25=8\times25+60 + 25=200+60 + 25 = 285 \) (thousand dollars)
Then, the cost at \( x = 5.25 \) is approximately \( C(5)+\Delta C=285 + 23=308 \) (thousand dollars)

Step1: Recall break - even point definition

The break - even point is when \( R(x)=C(x) \). Given \( R(x)=-x^2 + 61x + 55 \) and \( C(x)=8x^2+12x + 25 \)
Set \( -x^2 + 61x + 55=8x^2+12x + 25 \)

Step2: Rearrange the equation

Bring all terms to one side:
\( 8x^2+x^2+12x - 61x+25 - 55 = 0 \)
\( 9x^2-49x - 30 = 0 \)

Step3: Solve the quadratic equation

We use the quadratic formula \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \) for the quadratic equation \( ax^2+bx + c = 0 \). Here, \( a = 9 \), \( b=-49 \), \( c=-30 \)
First, calculate the discriminant \( D=b^2 - 4ac=(-49)^2-4\times9\times(-30)=2401 + 1080 = 3481 \)
\( \sqrt{D}=\sqrt{3481}=59 \)
Then, \( x=\frac{49\pm59}{18} \)
We have two solutions:
\( x_1=\frac{49 + 59}{18}=\frac{108}{18}=6 \)
\( x_2=\frac{49-59}{18}=\frac{-10}{18}=-\frac{5}{9} \)
Since the number of units \( x\geq0 \), we discard the negative solution.

Answer:

The marginal cost at \( x = 5 \) is 92 thousand dollars per unit.

Part (b)