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Question:- The scores given by three of the four professors are 45, 49, 52 for each of 2 students Alex and Dan.

What is the minimum possible final score of any student ?

My approach:-

Assuming the fourth score to be $x$.

Now $x$ has to be less than 52 otherwise , $Y$ (max deviation between the marks) as well as $A$ (Average would increase), but we need to minimize both as our final score is $A+(Y-X)$.

now if $x<45$, the Average would decrease but then $Y$ would increase, so for checking purpose I did put $x=0$ as a possibility, so final score turns out to be $\frac{0+45+49+52}{4} + 52-3 = 85.5$.

then I tried for $x=45$, for which I get $\frac{45+45+49+52}{4} + (7-0) = 54.75$.

now to check if I can minimize it further , I checked with $x=46$, $\frac{45+46+49+52}{4} + (7-1) = 48+6 =54$ (we achieve a more lower score)

then at $x=47$, $\frac{45+47+49+52}{4} + (7-2) = 53.25$.

now for $x=48$, $\frac{45+48+49+52}{4}+ (7-1) = 54.5$, so the score rises, can I say the minimum will be at $x=47$ and be equal to $53.25$.

Have I done it correctly? Any efficient way to handle this?

Tita
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Fin27
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1 Answers1

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For x in [0,42], X =3, A= 146+x/4 and Y = 52-x, so the final score is (146/4 + x/4) + (52-x)-3 which is minimized by 42. You can write piecewise functions like this for the remaining possible values. It is evident that large numbers (x > 55) increase both the value of Y and A, while X remains constant. So the range that needs checked is x between 42 and 52. Which is what you were doing manually. It appears to me that you have reached the correct solution.

Mark
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