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601-envthm - f x a = ax 2 − x So setting ∂ f x a ∂ x...

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T kh E qyhorsh T khruhp 1 Suppose we have a value function de fi ned by V ( a ) = Max or Min x f ( x, a ) , and f attains its max or min at a point where f/ x = 0 , then V ± ( a ) = f ( x ( a ) , a ) a . Why is this true? Let x ( a ) be the function that gives the optimal value of x for each value of a. Then V ( a ) = f ( x ( a ) , a ) . So V ± ( a ) = f ( x ( a ) , a ) x dx ( a ) da + f ( x ( a ) , a ) a . But the fi rst term on the RHS is zero because at the optimal value of x, f ( x ( a ) , a ) / x = 0 .
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Unformatted text preview: f ( x, a ) = ax 2 − x So setting ∂ f ( x, a ) / ∂ x = 0 we obtain x ( a ) = 1 2 a and V ( a ) = a ± 1 2 a ² 2 − 1 2 a = − 1 4 a and V ± ( a ) = 1 4 a 2 . But the envelope theorem allows us to get this result more easily. V ± ( a ) = ∂ f ( x, a ) ∂ a = ( x ( a )) 2 = 1 4 a 2 ....
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