Using lab $7$ with a three$-$day lead time $($all original values$),$ we will analyze six different scenarios to determine which would be the best scenario based on the total cost.

Scenario A $-$ Min is $3,500$ and Max is $5,500.$ All other values remain the same as lab $7$ original values. The total cost would be Blank $1$ Question $3.$

Scenario B $-$ Min is $3,500$ and Max is $6,000.$ All other values remain the same as lab $7$ original values. The total cost would be Blank $2$ Question $3.$

Scenario C $-$ Min is $4,000$ and Max is $5,500.$ All other values remain the same as lab $7$ original values. The total cost would be Blank $3$ Question $3.$

Scenario D $-$ Min is $4,000$ and Max is $6,000.$ All other values remain the same as lab $7$ original values. The total cost would be Blank $4$ Question $3.$

Scenario E $-$ Min is $4,500$ and Max is $5,500.$ All other values remain the same as lab $7$ original values. The total cost would be Blank $5$ Question $3.$

Scenario F $-$ Min is $4,500$ and Max is $6,000.$ All other values remain the same as lab $7$ original values. The total cost would be Blank $6$ Question $3.$

Based on the above scenarios, we know that scenario Blank $7$ Question $3$ would cost us the most, therefore we wouldn't want to use those min and max values. On the flip side, scenario Blank $8$ Question $3$ yields us the best combination of min and max resulting in the least cost.