513b we are given that lp 0 but by duality lp 2 πpw

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5.13.b We are given that ∂l/∂p< 0. But by duality, ∂l/∂p = 2 π/∂p∂w = 2 π/∂w∂p = ∂y/∂w. It follows that ∂y/∂w> 0. 5.14 Take a total derivative of the cost function to get: dc = n summationdisplay i =1 ∂c ∂w i dw i + ∂c ∂y dy. It follows that ∂c ∂y = dc n i =1 ∂c ∂w i dw i dy . Now substitute the first differences for the dy , dc , dw i terms and you’re done. 5.15 By the linearity of the function, we know we will use either x 1 , or a combination of x 2 and x 3 to produce y . By the properties of the Leontief function, we know that if we use x 2 and x 3 to produce y , we must use 3 units of both x 2 and x 3 to produce one unit of y . Thus, if the cost of using one unit of x 1 is less than the cost of using one unit of both x 2 and x 3 , then we will use only x 1 , and conversely. The conditional factor demands can be written as: x 1 = braceleftbigg 3 y if w 1 <w 2 + w 3 0 if w 1 >w 2 + w 3 x 2 = braceleftbigg 0 if w 1 <w 2 + w 3 3 y if w 1 >w 2 + w 3 x 3 = braceleftbigg 0 if w 1 <w 2 + w 3 3 y if w 1 >w 2 + w 3 if w 1 = w 2 + w 3 , then any bundle ( x 1 ,x 2 ,x 3 ) with x 2 = x 3 and x 1 + x 2 = 3 y (or x 1 + x 3 = 3 y ) minimizes cost. The cost function is c ( w,y ) = 3 y min( w 1 ,w 2 + w 3 ) . 5.16.a Homogeneous : c ( t w ,y ) = y 1 / 2 ( tw 1 tw 2 ) 3 / 4 = t 3 / 2 ( y 1 / 2 ( w 1 w 2 ) 3 / 4 ) = t 3 / 2 c ( w ,y ) No.
Ch. 5 COST FUNCTION 13 Monotone : ∂c ∂w 1 = 3 4 y 1 / 2 w 1 / 4 1 w 3 / 4 2 > 0 ∂c ∂w 2 = 3 4 y 1 / 2 w 3 / 4 1 w 1 / 4 2 > 0 Yes. Concave : Hessian = bracketleftBigg 3 16 y 1 / 2 w 5 / 4 1 w 3 / 4 2 9 16 y 1 / 2 w 1 / 4 1 w 1 / 4 2 9 16 y 1 / 2 w 1 / 4 1 w 1 / 4 2 3 16 y 1 / 2 w 3 / 4 1 w 5 / 4 2 bracketrightBigg | H 1 | < 0 | H 2 | = 9 256 yw 1 / 2 1 w 1 / 2 2 81 256 yw 1 / 2 1 w 1 / 2 2 = 72 256 y w 1 w 2 < 0 No Continuous : Yes 5.16.b Homogeneous : c ( t w ,y ) = y ( tw 1 + tw 1 tw 2 + tw 2 ) = ty ( w 1 + w 1 w 2 + w 2 ) = tc ( y, w ) Yes Monotone : ∂c ∂w 1 = y parenleftbigg 1 + 1 2 radicalbigg w 2 w 1 parenrightbigg > 0 ∂c ∂w 2 = y parenleftbigg 1 + 1 2 radicalbigg w 1 w 2 parenrightbigg > 0 Yes Concave : H = bracketleftBigg 1 4 yw 1 / 2 2 w 3 / 2 1 1 4 yw 1 / 2 2 w 1 / 2 1 1 4 yw 1 / 2 2 w 1 / 2 1 1 4 yw 3 / 2 2 w 1 / 2 1 bracketrightBigg | H 1 | < 0 | H 2 | = 1 16 yw 1 2 w 1 1 1 16 yw 1 2 w 1 1 = 0 Yes Continuous : Yes
14 ANSWERS Production Function : x 1 ( w ,y ) = y parenleftbigg 1 + 1 2 radicalbigg w 2 w 1 parenrightbigg (1) x 2 ( w ,y ) = y parenleftbigg 1 + 1 2 radicalbigg w 1 w 2 parenrightbigg (2) Rearranging these equations: x 1 y = y 2 radicalbigg w 2 w 1 (1 ) x 2 y = y 2 radicalbigg w 1 w 2 (2 ) Multiply (1 ) and (2 ): ( x 1 y )( x 2 y ) = y 2 4 . This is a quadratic equation which gives y = 2 3 ( x 2 + x 1 ) ± 2 3 radicalbig x 2 1 + x 2 2 + 2 x 1 x 2 . 5.16.c Homogeneous : c ( t w ,y ) = y ( tw 1 e tw 1 + tw 2 ) = ty ( w 1 e tw 1 + w 2 ) negationslash = tc ( w ,y ) No Monotone: ∂c ∂w 1 = y ( w 1 e w 1 + e w 1 ) = ye w 1 (1 w 1 ) This is positive only if w 1 < 1.

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