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IB Math AA HL IA - 7

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IB Math AA HL IA Grade: 7 Topic: Utility Maximization: How many cups of tea and coffee to have within a budget constraint? Lagrangian multiplier method (Calculus) was used.

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Uploaded on
February 23, 2024
Number of pages
15
Written in
2020/2021
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Essay
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Ma he a c IA

U Ma a :H a c f ea
a d c ffee ha e h a b dge c a ?

W dc : 4322

, I d ci
Thi S e,I e e i e h ida bac i h e a d beca e addic ed af d de i e
a ica i d e he a ai abi i f ai ch ice a d c e ie ce. S ei e ,I d de a
f h ee ea e e h gh he e e e e i g edie b gh i he ef ige a f e
c . Be ide , I e e de d i de e e e da af e ea . O e da I e e ade 7 de
ha a ica i fi ach.
N de a aged a he c ide ab e a f e I e ha a ica i .
She f d ha I ed a 1000 (Chi e e c e c a ,e i ae a i ae 128 a
ha i e) he ga e e a he a f h ida i e ee . I ha e ad i ha I a e ib e a
a agi g fi a ce . The ef e, dd ha f e.
She a b dge c ai a ie a df d. I a ic a , he a a ge ed a he fac ha I
de he f i ge e i ed i e e da . A d I a ha e ad i ha he e di ae
i deed e e i e beca e I ca fi d a e ea f ch ha c ai e gh de ici f da d
ae ch chea e ha he . A a e , he d a e e da 400 a h
he d i . She c d chec ha a a i e beca e f d de i e a ica i i i ed i h he
c edi ca d a d e e a ac i I ade i be ec ded i de ai acce ib e f b h acc
a d he .
M fa i e c ffee d i c 26.5 e c ( 25 + 1.5 ac age fee) a d fa i e ea d i
c 28.5 e c ( 27 + 1.5 ac age fee). Be ide , I d e j he e a - I fee i e I i e
he ea d i e ha he c ffee d i . Gi e he fac ha I ha e a b dge f 400 a h, I
de h a c f ea a d c ffee I h db i a hf e ha e he g ea e
a i fac i f d i i g he hich i he ai f i e iga i .
Whe I hi ab ha i g g ea e a i fac i , I i edia e ea i e ha be i e a ed a
ec ic c ce ca ed i i . Af e d i g, I a de a d be i ac
c ai ed i i ai e i i i a iab e ca c ha eed be ed i g he Lag a ge
i ie e h d.


E planation of the method of lagrange m ltiplier:
The e h d f ag a ge i ie i a ef fi d e e e a e fc ai ed f ci ,
i.e., f ci i hi c ai . The e h d f ag a ge i ie ae :
F diffe e iab e f ci f ( , , ) a d ( , , ), he ca a i a d i i a e ff
bjec he c ai ( , , ) = 0 ca be f d b fi di g he a e f , , a d ha
ai f he e a i :
∇f = ∇g and g ( , , ) = 0, ∈ ℝ+




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