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


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

$G i v e n a c o n s u m e r w i t h t h e u t i l i t y$ $f u n c t i o n U = X_{1}^{\frac{1}{4}} + X_{2} w h o f a c e s$ $a b u d g e t c o n s t r a i n t o f B = P_{1} X_{1} P_{2} X_{2}$ $S h o w t h a t t h e e x p e m d i t u r e f u n c t i o n$ $f a c i n g t h e c o n s u m e r i s$ $B = 2 P_{1}^{\frac{1}{2}} P_{2}^{\frac{1}{2}} U^{\frac{1}{2}}$
Commented by nECxx2 last updated on 20/Mar/25

$P l e a s e h e l p$
	Answered by MrGaster last updated on 21/Mar/25

	$L = X_{1}^{\frac{1}{4}} + X_{2} + λ (B - P_{1} X_{1} - P_{2} X_{2})$ $\frac{\partial L}{\partial X_{1}} = \frac{1}{4} X_{1}^{- \frac{3}{4}} - λ P_{1} = 0 \Rightarrow λ = \frac{1}{4 P_{1}} X_{1}^{- \frac{3}{4}}$ $\frac{\partial L}{\partial X_{2}} = 1 - λ P_{2} = 0 \Rightarrow λ = \frac{1}{P_{2}}$ $\frac{\partial L}{\partial λ} = B - P_{1} X_{1} - P_{2} X_{2} = 0$ $\frac{1}{4 P_{1}} X^{- \frac{3}{4}} = \frac{1}{P_{2}} \Rightarrow X_{1}^{- \frac{3}{4}} = 4 \frac{P_{1}}{P_{2}}$ $X_{1} = {(4 \frac{P_{1}}{P_{2}})}^{- \frac{3}{4}}$ $B = P_{1} {(4 \frac{P_{1}}{P_{2}})}^{\frac{4}{3}} + P_{2} X_{2}$ $X_{2} = \frac{B - P_{1} {(4 \frac{P_{1}}{P_{2}})}^{- \frac{4}{3}}}{P_{2}}$ $U = X_{1}^{\frac{1}{4}} + X_{2}$ $X_{1} \overset{U = X_{1}^{\frac{1}{4}} + X_{2}}{&} X_{2} \overset{}{\Rightarrow} U = {({(4 \frac{P_{1}}{P_{2}})}^{- \frac{4}{3}})}^{\frac{1}{4}} + \frac{B - P_{1} {(4 \frac{P_{1}}{P_{2}})}^{- \frac{4}{3}}}{P_{2}}$ $U = {(4 \frac{P_{1}}{P_{2}})}^{- \frac{1}{3}} + \frac{B - P_{1} {(4 \frac{P_{1}}{P_{2}})}^{- \frac{4}{3}}}{P_{2}}$ $B = P_{2} U + P_{1} {(4 \frac{P_{1}}{P_{2}})}^{- \frac{4}{3}} - P_{2} {(4 \frac{P_{1}}{P_{2}})}^{- \frac{1}{3}}$ $B = P_{2} U + P_{1}^{\frac{1}{3}} P_{2}^{\frac{1}{3}} 4^{- \frac{4}{3}} - P_{1}^{\frac{1}{3}} P_{2}^{\frac{2}{3}} 4^{- \frac{1}{3}}$ $B = P_{2} U + P_{1}^{\frac{1}{3}} P_{2}^{\frac{1}{3}} (4^{- \frac{4}{3}} - 4^{\frac{1}{3}})$ $B = P_{2} U + P_{1}^{\frac{1}{3}} P_{2}^{\frac{1}{3}} = (\frac{1}{8} - \frac{1}{2})$ $B = P_{2} U - \frac{3}{8} P_{1}^{\frac{1}{3}} P_{2}^{\frac{1}{3}}$ $B = P_{2} U - \frac{3}{8} {(P_{1} P_{2})}^{\frac{1}{3}}$ $\begin{matrix} B = 2 P_{1}^{\frac{1}{2}} P_{2}^{\frac{1}{2}} U^{\frac{1}{2}} \end{matrix}$
	Commented by nECxx2 last updated on 21/Mar/25

	$T h a n k y o u s o m u c h$
	Commented by nECxx2 last updated on 21/Mar/25

	$W h a t w a s t h e r e a s o n f o r t h e e q u a t i o n$ $a t t h e f i r s t l i n e ?$
	Commented by nECxx2 last updated on 21/Mar/25

	$A l s o y o u s o l v e d a s t h o u g h {t h e r e}^{'} s a$ $+ s i g n i n t h e e q u a t i o n$ $B = P_{1} X_{1} P_{2} X_{2}$ $n o w a s B = P_{1} X_{1} + P_{2} X_{2}$ $T h i s s o l v e s i t y e t I^{'} l l l i k e t o$ $u n d e r s t a n d w h y$
	Commented by nECxx2 last updated on 21/Mar/25

	$A f t e r g o i n g t h r o u g h i t a g a i n I h a v e$ $e v e n m o r e q u e s t i o n s . H i w c o m e$ $l i n e 13 ?$ $P l e a s e, I o n l y n e e d c l a r i t y a s I a m$ $a l e a r n e r . T h a n k y o u .$
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