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

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

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

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

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

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

$$424,627$

$$246,492$

$$301,993$