Math Problem Statement

Solve from 9th question explaining every single calculation

Solve from 9th question explaining every single calculation.

Solution

Let’s solve the questions starting from Question 9. I'll break down each step of the calculations and provide clear explanations.


Question 9:

The cost function for xx units is given by:
C(x)=ax3+bx2cx+d,where a>0,b<0,c>0C(x) = ax^3 + bx^2 - cx + d, \quad \text{where } a > 0, b < 0, c > 0
We are asked to show that average variable cost (AVC) and marginal cost (MC) intersect at the minimum of the average variable cost.


Step 1: Definitions

  1. Total Cost (TC): Given as C(x)=ax3+bx2cx+dC(x) = ax^3 + bx^2 - cx + d.
    Here, dd is the fixed cost (constant term).

  2. Variable Cost (VC): Variable cost is the part of total cost dependent on xx, so:
    VC(x)=ax3+bx2cxVC(x) = ax^3 + bx^2 - cx

  3. Average Variable Cost (AVC):
    Average Variable Cost is:
    AVC(x)=VC(x)x=ax3+bx2cxx=ax2+bxcAVC(x) = \frac{VC(x)}{x} = \frac{ax^3 + bx^2 - cx}{x} = ax^2 + bx - c

  4. Marginal Cost (MC):
    Marginal Cost is the derivative of total cost C(x)C(x) with respect to xx:
    MC(x)=ddx[ax3+bx2cx+d]=3ax2+2bxcMC(x) = \frac{d}{dx} \big[ax^3 + bx^2 - cx + d\big] = 3ax^2 + 2bx - c


Step 2: Finding Intersection Point

To find the intersection of AVC(x)AVC(x) and MC(x)MC(x), set AVC(x)=MC(x)AVC(x) = MC(x):
ax2+bxc=3ax2+2bxcax^2 + bx - c = 3ax^2 + 2bx - c

Simplify by subtracting ax2+bxcax^2 + bx - c from both sides:
0=2ax2+bx0 = 2ax^2 + bx

Factorize:
x(2ax+b)=0x(2ax + b) = 0

Thus, x=0x = 0 or x=b2ax = -\frac{b}{2a}.
Since x=0x = 0 is not feasible for production, the intersection occurs at:
x=b2ax = -\frac{b}{2a}


Step 3: Verify Minimum Point

To verify that this is the minimum of AVC(x)AVC(x), compute the derivative of AVC(x)AVC(x):
AVC(x)=ax2+bxcAVC(x) = ax^2 + bx - c ddx[AVC(x)]=2ax+b\frac{d}{dx} [AVC(x)] = 2ax + b

Set ddx[AVC(x)]=0\frac{d}{dx} [AVC(x)] = 0 for critical points:
2ax+b=0    x=b2a2ax + b = 0 \implies x = -\frac{b}{2a}

Thus, AVC(x)AVC(x) is minimum at x=b2ax = -\frac{b}{2a}, which is the same point where AVC(x)=MC(x)AVC(x) = MC(x).


Conclusion for Question 9:

The average variable cost (AVC) and marginal cost (MC) curves intersect at the minimum of the average variable cost curve.


Would you like to proceed with Question 10 or review any steps in this solution? Let me know!

Here are five related questions you might consider:

  1. What is the significance of the intersection of AVC and MC?
  2. How does fixed cost impact the cost functions?
  3. Why does AVC(x)AVC(x) reach a minimum?
  4. How does the shape of AVC(x)AVC(x) change for different values of a,b,ca, b, c?
  5. Can AVC and MC fail to intersect for certain cost functions?

Tip: Always verify minimum/maximum points by checking the derivative's sign or the second derivative.

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Math Problem Analysis

Mathematical Concepts

Calculus
Cost Analysis
Optimization

Formulas

Total Cost: C(x) = ax^3 + bx^2 - cx + d
Variable Cost: VC(x) = ax^3 + bx^2 - cx
Average Variable Cost: AVC(x) = VC(x)/x = ax^2 + bx - c
Marginal Cost: MC(x) = d(C(x))/dx = 3ax^2 + 2bx - c

Theorems

Intersection of AVC and MC at minimum AVC point

Suitable Grade Level

Grades 11-12 or introductory college-level calculus