During the next two months an automobile manufacturer must meet (on time) the following demands for trucks and cars: month 1, 400 trucks and 800 cars; month 2, 300 trucks and 300 cars. During each month at most 1000 vehicles can be produced. Each truck uses two tons of steel, and each car uses one ton of steel. During month 1, steel costs $700 per ton; during month 2, steel is projected to cost $800 per ton. At most 2500 tons of steel can be purchased each month. (Steel can be used only during the month in which it is purchased.) At the beginning of month 1, 100 trucks and 200 cars are in the inventory. At the end of each month, a holding cost of $200 per vehicle is assessed. Each car gets 35 miles per gallon (mpg), and each truck gets 15 mpg. During each month, the vehicles produced by the company must average at least 23 mpg. Required:
a. Determine how to meet the demand and mileage requirements at minimum total cost.
b. Explain intuitively what happens when the requirement is greater than 17 mpg.

a. solver found a solution 600C₁ + 400T₁ + 300C₂ + 200 T₂

the company should produce 600 cars and 400 trucks during month 1, and it should produce 300 cars and 200 trucks during month 2

b. here it doesn't really make a lot of difference since a car's mpg is 35 and a truck's is 15. The average between them is 25. You would need to produce more than 9 trucks per car (at least 10) in order for the average to be lower than 17 mpg.

Explanation:

minimization equation = 700S₁ + 800S₂ + 200HC₁ + 200HT₁ + 200HC₂ + 200HC₂

where:  

C₁ = cars produced during month 1  

T₁ = trucks produced during month 1

C₂ = cars produced during month 1

T₂ = trucks produced during month 2  

S₁ = steel used during month 1

S₂ = steel used during month 2

HC₁ = holding cost for a car on month 1

HT₁ = holding cost for a truck on month 1

HC₂ = holding cost for a car on month 2

HT₂ = holding cost for a truck on month 1  

constraints:  

C₁ + T₁ ≤ 1000    

C₂ + T₂ ≤ 1000    

-S₁ + C₁ + 2T₁ = 0    

-S₂ + C₂ + 2T₂ = 0

-HC₁ + C₁ ≥ 600

-HT₁ + T₁ ≥ 300

HC₁ - HC₂ + C₂ ≥ 300

HT₁ - HT₂ + T₂ ≥ 300

4C₁ - 6T₁ ≥ 0

4C₂ - 6T₂ ≥ 0  

S₁ ≤ 2500  

S₂ ≤ 2500

solver found a solution 600C₁ + 400T₁ + 300C₂ + 200 T₂

answer: marginal utility of the fourth brownie is 0.

explanation: marginal utility is the addtional utility gained from the consumption of one more unit of a good. it is given by,

so, marginal utility at the 4th unit will be given by

since, the values are not given here, you can simply find marginal utility at fourth unit by subtracting total utility at 3rd unit from total utility at 4th unit.