there are three exams in your linear algebra class, and you theorize that your s
ID: 2943531 • Letter: T
Question
there are three exams in your linear algebra class, and you theorize that your score in each exam (out of 100) will be numerically equal to the number of hours you study for that exam. The three exams count 20%,30%, and 50%, respectively, toward the final grade. If your (modest) goal is to score 76% in the course, how many hours a, b, and c should you study for each of the three exams to minimize quantity a2+b2+c2? This quadratic model reflects he fact that it may be four times as painful to study for 10 hours than for just 5 hours.
Explanation / Answer
there are three exams in your linear algebra class, and you theorize that your score in each exam (out of 100) will be numerically equal to the number of hours you study for that exam. The three exams count 20%,30%, and 50%, respectively, toward the final grade. If your (modest) goal is to score 76% in the course, how many hours a, b, and c should you study for each of the three exams to minimize quantity a2+b2+c2? This quadratic model reflects he fact that it may be four times as painful to study for 10 hours than for just 5 hours
EXPECTED MARKS = K*HRS PUT IN FOR STUDY
HENCE
MARKS IN EACH TOPIC ARE KA,KB,KC.
TAKING IN TO ACCOUNT THEIR WEIGHTAGE AS GIVEN
GRADE OBTAINED = 0.2*KA+0.3*KB*0.3*KC=76 AS NEEDED.....
SO WE HAVE THE CONSTRAINT AS ..
K[0.2A+0.3B+0.5C]=76.......................................1
FUNCTION TO BE MINIMIZED IS
F=A*A+B*B+C*C..................................2
DIFFERENTIATING 1 AND 2 WE GET
K[0.2DA+0.3DB+0.5DC]=0
0.2DA+0.3DB+0.5DC=0...........................3
2ADA+2BDB+2CDC=0......FOR OPTIMUM
ADA+BDB+CDC=0............................4
EQN.4+L*EQN.3 GIVES US
[A+0.2L]DA+[B+0.3L]DB+[C+0.5L]DC=0
A+0.2L=0................-L=5A....................5
B+0.3L=0...............-L=10B/3................6
C+0.5L=0.............-L=2C.............................7
ELIMINATING L WE GET
5A=10B/3=2C
B=1.5A.............................8
C=2.5A.................................9
PUTTING 8 AND 9 IN 1 WE GET
K[0.2A+0.3*1.5A+0.5*2.5A]=76
KA[0.2+0.45+1.25]=76
A=76/[1.9K]
WELL THE ANSWER IN NUMBERS DEPEND ON K ...
IF WE TAKE K=1 FOR EXAMPLE [THAT IS YOU GET AS MANY MARKS AS THE HOURS YOU PUT IN ..]
A=76/1.9=40 HRS
B=1.5*40=60 HRS
C=2.5*40=100 HRS.
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