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A difficult math course
Solution: Let the total time be y hours.

y 1=4000/6x

y2=3000/3(2 16-x)

Y is the larger value of y 1 and y2.

Y 1 > when x≤86; y2,y = y 1; Y 1

When x≤86, y decreases with x, so when x=86, y takes the minimum value 1000/ 129.

When x≥87, y increases with x, so when x=87, y takes the minimum value 1000/ 129.

To sum up, y takes the minimum value when x=86 or 87.

So 86 people were assigned to process G device, 130 people to process H device; Or assign 87 people to process G device, 129 people to process H device.

This is also called "math difficulty"?