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Question Number 217769 by nECxx2 last updated on 20/Mar/25

Given a consumer with the utility function U = X_1 ^(1/4) + X_2 who faces a budget constraint of B=P_1 X_1 P_2 X_2 Show that the expemditure function facing the consumer is B = 2P_1 ^(1/2) P_2 ^(1/2) U^(1/2)

$${Given}\:{a}\:{consumer}\:{with}\:{the}\:{utility} \\ $$$${function}\:{U}\:=\:{X}_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{4}}} +\:{X}_{\mathrm{2}} \:{who}\:{faces} \\ $$$${a}\:{budget}\:{constraint}\:{of}\:{B}={P}_{\mathrm{1}} {X}_{\mathrm{1}} {P}_{\mathrm{2}} {X}_{\mathrm{2}} \\ $$$${Show}\:{that}\:{the}\:{expemditure}\:{function} \\ $$$${facing}\:{the}\:{consumer}\:{is} \\ $$$${B}\:=\:\mathrm{2}{P}_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {P}_{\mathrm{2}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {U}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$

Commented by nECxx2 last updated on 20/Mar/25

Please help

$${Please}\:{help} \\ $$

Answered by MrGaster last updated on 21/Mar/25

L=X_1 ^(1/4) +X_2 +λ(B−P_1 X_1 −P_2 X_2 ) (∂L/∂X_1 )=(1/4)X_1 ^(−(3/4)) −λP_1 =0⇒λ=(1/(4P_1 ))X_1 ^(−(3/4)) (∂L/∂X_2 )=1−λP_2 =0⇒λ=(1/P_2 ) (∂L/∂λ)=B−P_1 X_1 −P_2 X_2 =0 (1/(4P_1 ))X^(−(3/4)) =(1/P_2 )⇒X_1 ^(−(3/4)) =4(P_1 /P_2 ) X_1 =(4(P_1 /P_2 ))^(−(3/4)) B=P_1 (4(P_1 /P_2 ))^(4/3) +P_2 X_2 X_2 =((B−P_1 (4(P_1 /P_2 ))^(−(4/3)) )/P_2 ) U=X_1 ^(1/4) +X_2 X_1 &^(U=X_1 ^(1/4) +X_2 ) X_2 ⇒^( ) U=((4(P_1 /P_2 ))^(−(4/3)) )^(1/4) +((B−P_1 (4(P_1 /P_2 ))^(−(4/3)) )/P_2 ) U=(4(P_1 /P_2 ))^(−(1/3)) +((B−P_1 (4(P_1 /P_2 ))^(−(4/3)) )/P_2 ) B=P_2 U+P_1 (4(P_1 /P_2 ))^(−(4/3)) −P_2 (4(P_1 /P_2 ))^(−(1/3)) B=P_2 U+P_1 ^(1/3) P_2 ^(1/3) 4^(−(4/3)) −P_1 ^(1/3) P_2 ^(2/3) 4^(−(1/3)) B=P_2 U+P_1 ^(1/3) P_2 ^(1/3) (4^(−(4/3)) −4^(1/3) ) B=P_2 U+P_1 ^(1/3) P_2 ^(1/3) =((1/8)−(1/2)) B=P_2 U−(3/8)P_1 ^(1/3) P_2 ^(1/3) B=P_2 U−(3/8)(P_1 P_2 )^(1/3) determinant (((B=2P_1 ^(1/2) P_2 ^(1/2) U^(1/2) )))

$$\mathcal{L}={X}_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{4}}} +{X}_{\mathrm{2}} +\lambda\left({B}−{P}_{\mathrm{1}} {X}_{\mathrm{1}} −{P}_{\mathrm{2}} {X}_{\mathrm{2}} \right) \\ $$$$\frac{\partial\mathcal{L}}{\partial{X}_{\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{4}}{X}_{\mathrm{1}} ^{−\frac{\mathrm{3}}{\mathrm{4}}} −\lambda{P}_{\mathrm{1}} =\mathrm{0}\Rightarrow\lambda=\frac{\mathrm{1}}{\mathrm{4}{P}_{\mathrm{1}} }{X}_{\mathrm{1}} ^{−\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$\frac{\partial\mathcal{L}}{\partial{X}_{\mathrm{2}} }=\mathrm{1}−\lambda{P}_{\mathrm{2}} =\mathrm{0}\Rightarrow\lambda=\frac{\mathrm{1}}{{P}_{\mathrm{2}} } \\ $$$$\frac{\partial\mathcal{L}}{\partial\lambda}={B}−{P}_{\mathrm{1}} {X}_{\mathrm{1}} −{P}_{\mathrm{2}} {X}_{\mathrm{2}} =\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}{P}_{\mathrm{1}} }{X}^{−\frac{\mathrm{3}}{\mathrm{4}}} =\frac{\mathrm{1}}{{P}_{\mathrm{2}} }\Rightarrow{X}_{\mathrm{1}} ^{−\frac{\mathrm{3}}{\mathrm{4}}} =\mathrm{4}\frac{{P}_{\mathrm{1}} }{{P}_{\mathrm{2}} } \\ $$$${X}_{\mathrm{1}} =\left(\mathrm{4}\frac{{P}_{\mathrm{1}} }{{P}_{\mathrm{2}} }\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$${B}={P}_{\mathrm{1}} \left(\mathrm{4}\frac{{P}_{\mathrm{1}} }{{P}_{\mathrm{2}} }\right)^{\frac{\mathrm{4}}{\mathrm{3}}} +{P}_{\mathrm{2}} {X}_{\mathrm{2}} \\ $$$${X}_{\mathrm{2}} =\frac{{B}−{P}_{\mathrm{1}} \left(\mathrm{4}\frac{{P}_{\mathrm{1}} }{{P}_{\mathrm{2}} }\right)^{−\frac{\mathrm{4}}{\mathrm{3}}} }{{P}_{\mathrm{2}} } \\ $$$${U}={X}_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{4}}} +{X}_{\mathrm{2}} \\ $$$${X}_{\mathrm{1}} \overset{{U}={X}_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{4}}} +{X}_{\mathrm{2}} } {\&}{X}_{\mathrm{2}} \overset{\: } {\Rightarrow}{U}=\left(\left(\mathrm{4}\frac{{P}_{\mathrm{1}} }{{P}_{\mathrm{2}} }\right)^{−\frac{\mathrm{4}}{\mathrm{3}}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} +\frac{{B}−{P}_{\mathrm{1}} \left(\mathrm{4}\frac{{P}_{\mathrm{1}} }{{P}_{\mathrm{2}} }\right)^{−\frac{\mathrm{4}}{\mathrm{3}}} }{{P}_{\mathrm{2}} } \\ $$$${U}=\left(\mathrm{4}\frac{{P}_{\mathrm{1}} }{{P}_{\mathrm{2}} }\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} +\frac{{B}−{P}_{\mathrm{1}} \left(\mathrm{4}\frac{{P}_{\mathrm{1}} }{{P}_{\mathrm{2}} }\right)^{−\frac{\mathrm{4}}{\mathrm{3}}} }{{P}_{\mathrm{2}} } \\ $$$${B}={P}_{\mathrm{2}} {U}+{P}_{\mathrm{1}} \left(\mathrm{4}\frac{{P}_{\mathrm{1}} }{{P}_{\mathrm{2}} }\right)^{−\frac{\mathrm{4}}{\mathrm{3}}} −{P}_{\mathrm{2}} \left(\mathrm{4}\frac{{P}_{\mathrm{1}} }{{P}_{\mathrm{2}} }\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${B}={P}_{\mathrm{2}} {U}+{P}_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{3}}} {P}_{\mathrm{2}} ^{\frac{\mathrm{1}}{\mathrm{3}}} \mathrm{4}^{−\frac{\mathrm{4}}{\mathrm{3}}} −{P}_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{3}}} {P}_{\mathrm{2}} ^{\frac{\mathrm{2}}{\mathrm{3}}} \mathrm{4}^{−\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${B}={P}_{\mathrm{2}} {U}+{P}_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{3}}} {P}_{\mathrm{2}} ^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{4}^{−\frac{\mathrm{4}}{\mathrm{3}}} −\mathrm{4}^{\frac{\mathrm{1}}{\mathrm{3}}} \right) \\ $$$${B}={P}_{\mathrm{2}} {U}+{P}_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{3}}} {P}_{\mathrm{2}} ^{\frac{\mathrm{1}}{\mathrm{3}}} =\left(\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${B}={P}_{\mathrm{2}} {U}−\frac{\mathrm{3}}{\mathrm{8}}{P}_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{3}}} {P}_{\mathrm{2}} ^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$${B}={P}_{\mathrm{2}} {U}−\frac{\mathrm{3}}{\mathrm{8}}\left({P}_{\mathrm{1}} {P}_{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\begin{array}{|c|}{{B}=\mathrm{2}{P}_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {P}_{\mathrm{2}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {U}^{\frac{\mathrm{1}}{\mathrm{2}}} }\\\hline\end{array} \\ $$

Commented by nECxx2 last updated on 21/Mar/25

Thank you so much

$${Thank}\:{you}\:{so}\:{much} \\ $$

Commented by nECxx2 last updated on 21/Mar/25

What was the reason for the equation at the first line?

$${What}\:{was}\:{the}\:{reason}\:{for}\:{the}\:{equation} \\ $$$${at}\:{the}\:{first}\:{line}? \\ $$

Commented by nECxx2 last updated on 21/Mar/25

Also you solved as though there′s a + sign in the equation B = P_1 X_1 P_2 X_2 now as B=P_1 X_1 +P_2 X_2 This solves it yet I′ll like to understand why

$${Also}\:{you}\:{solved}\:{as}\:{though}\:{there}'{s}\:{a} \\ $$$$+\:{sign}\:{in}\:{the}\:{equation} \\ $$$${B}\:=\:{P}_{\mathrm{1}} {X}_{\mathrm{1}} {P}_{\mathrm{2}} {X}_{\mathrm{2}} \\ $$$${now}\:{as}\:{B}={P}_{\mathrm{1}} {X}_{\mathrm{1}} +{P}_{\mathrm{2}} {X}_{\mathrm{2}} \\ $$$${This}\:{solves}\:{it}\:{yet}\:{I}'{ll}\:{like}\:{to} \\ $$$${understand}\:{why} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by nECxx2 last updated on 21/Mar/25

After going through it again I have even more questions. Hiw come line 13? Please, I only need clarity as I am a learner. Thank you.

$${After}\:{going}\:{through}\:{it}\:{again}\:{I}\:{have} \\ $$$${even}\:{more}\:{questions}.\:{Hiw}\:{come} \\ $$$${line}\:\mathrm{13}? \\ $$$${Please},\:{I}\:{only}\:{need}\:{clarity}\:{as}\:{I}\:{am} \\ $$$${a}\:{learner}.\:{Thank}\:{you}. \\ $$

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