Suppose you want to buy a new car and are trying to choose between two models: Model A: costs $16,500 and its gas mileage is 25 miles per gallon and its insurance is $250 per year. Model B: costs $24,500 and its gas mileage is 40 miles per gallon and its insurance is $450 per year. If you drive approximately 40,000 miles per year and the gas costs $3 per gallon:
1. Find a formula for the total cost of owning Model A where the number of years you own the car is represented by x.
2. Find a formula for the total cost of owning Model B where the number of years is the independent variable.
3. Find the total cost for each model for the first five years. If you plan to keep the car for 4 years, which model is more economical?
4. Find the number of years in which the total cost to keep the two cars will be the same.
5. Identify the number of months where neither car holds a cost of ownership advantage.
6. What effect would the cost of gas doubling have on cost of ownership?
7. If you can sell neither car for 40% of its value at any time, how does the analysis change?

6. The cost of ownership of Model A increases more than Model B, as it is less gas efficient. The break-even point for x is reduced from x=5 to x=2.35.

7. The fixed cost are reduced by a 40%, so the variable cost, the ones that depend on time of ownership, are increased in importance.

Step-by-step explanation:

We can express the cost of ownership as the sum of the purchase cost, gas cost and insurance cost.

1. For model A we have:

[tex]\text{Cost of ownership}=\text{Purchase cost}+\text{Gas cost}+\text{Insurance cost}\\\\\text{Cost of ownership}=\$16,500+3 \dfrac{\$}{gal}\cdot\dfrac{1\,gal}{25\,miles}\cdot \dfrac{40,000\,miles}{year}\cdot x+\$250\cdot x\\\\\\\text{Cost of ownership}=\$16,500+\$4,800x+\$250x\\\\\\\text{Cost of ownership}=$16,500+\$5,050x[/tex]

2. For model B we have:

[tex]\text{Cost of ownership}=\text{Purchase cost}+\text{Gas cost}+\text{Insurance cost}\\\\\text{Cost of ownership}=\$24,500+3 \dfrac{\$}{gal}\cdot\dfrac{1\,gal}{40\,miles}\cdot \dfrac{40,000\,miles}{year}\cdot x+\$450\cdot x\\\\\\\text{Cost of ownership}=\$24,500+\$3,000x+\$450x\\\\\\\text{Cost of ownership}=$24,500+\$3,450x[/tex]

3. If x=5, the costs for each car are:

[tex]\text{CoOwn A}=16,500+5,050\cdot(5)=16,500+25,250=41,750\\\\\\\text{CoOwn B}=24,500+3,450\cdot(5)=24,500+17,250=41,750[/tex]

5 years is the break-even point for the cost of ownership between these two cars.

If you plan to keep the car for 4 years, the costs are:

[tex]\text{CoOwn A}=16,500+5,050\cdot(4)=16,500+20,200=36,700\\\\\\\text{CoOwn B}=24,500+3,450\cdot(4)=24,500+13,800=38,300[/tex]

For a 4 year period ownership, the model A is more economical ($36,700).

4. This happens for 1 year, the fifth year, in which the two models have the same cost of ownership.

5. At the 5th year, the cost for both models are the same.

Then, this corresponds to the months 4*12+1=48+1=49 and 5*12+1=61.

6. If the cost of gas doubles, the cost of ownership would rise for both model, but more for the Model A, which is less gas efficient and hence has a higher gas cost.

Model A

[tex]\text{Cost of ownership}=\text{Purchase cost}+\text{Gas cost}+\text{Insurance cost}\\\\\text{Cost of ownership}=\$16,500+6 \dfrac{\$}{gal}\cdot\dfrac{1\,gal}{25\,miles}\cdot \dfrac{40,000\,miles}{year}\cdot x+\$250\cdot x\\\\\\\text{Cost of ownership}=\$16,500+\$9,600x+\$250x\\\\\\\text{Cost of ownership}=$16,500+\$9,850x[/tex]

Model B

[tex]\text{Cost of ownership}=\text{Purchase cost}+\text{Gas cost}+\text{Insurance cost}\\\\\text{Cost of ownership}=\$24,500+6 \dfrac{\$}{gal}\cdot\dfrac{1\,gal}{40\,miles}\cdot \dfrac{40,000\,miles}{year}\cdot x+\$450\cdot x\\\\\\\text{Cost of ownership}=\$24,500+\$6,000x+\$450x\\\\\\\text{Cost of ownership}=$24,500+\$6,450x[/tex]

The breakeven point goes from x=5 (for $3 per gallon) to x=2.35 (for $6 per gallon).

[tex]16,500+9,850x=24,500+6,450x\\\\(9,850-6,450)x=24,500-16,500\\\\x=8,000/3400=2.35[/tex]

7. If we can sell any car for 40% of its value at any time, the cost of ownership becames:

Model A:

[tex]\text{Cost of ownership}=16,500+5,050x-0.4\cdot16,500\\\\\text{Cost of ownership}=9,900+5,050x[/tex]

Model B

[tex]\text{Cost of ownership}=24,500+3,450x-0.4\cdot24,500\\\\\text{Cost of ownership}=14,700+3,450x[/tex]

The fixed costs are lowered by 40%, so the variable costs (the ones that depend on time) became more important.