# Ephraim 44

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Calculus: Lagrange Multipliers
1.
a)
Here,
 
    
Getting the critical points,
 
 
Hence,


Thus, the critical point is at

This has to be the location of the minimum, as there are no other
critical points. [You can also do completing the squares and see that those perfect squares have a
minimum value of 0. Hence, this function only has a minimum.]
Hence, plugging this to h(x,y),


 
 



 
  




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b)
Maximizing
getting the critical points,
  
Hence, the critical value is


[Note: This has to be the answer as there are no other critical points to choose from.]
Hence, the maximum value of





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Calculus: Lagrange Multipliers 1. a) Here, ℎ(𝑥, 𝑦) = 𝑥 2 + 𝑦 2 − 𝜆(2𝑥 + 4𝑦 − 20) Getting the critical points, ℎ𝑥 = 0 = 2𝑥 − 2𝜆 ℎ𝑦 = 0 = 2𝑦 − 4𝜆 Hence, 𝑥= 𝜆 𝑦 = 2𝜆 Thus, the critical point is at (𝜆, 2𝜆). This has to be the location of the minimum, as there are no other critical points. [You can also do completing the squares and see that those perfect squares have a minimum value of 0. Hence, this function only has a minimum.] Hence, plugging this to h(x,y), 𝑚(𝜆) = 𝑓(𝜆, 2𝜆) = 𝜆2 + (2𝜆)2 − 𝜆(2(𝜆) + 4(2𝜆) − 20) = 5𝜆2 − 𝜆(10𝜆 − 20) 𝒎(𝝀) = −𝟓𝝀𝟐 + 𝟐𝟎𝝀 = −𝟓𝑳𝟐 + 𝟐𝟎𝑳 [𝑨𝑵𝑺𝑾𝑬𝑹] b) Maximizing 𝑚(𝜆), getting the critical points, 𝑚𝜆 = −10𝜆 + 20 = 0 Hence, the critical value is 𝝀 = 𝟐 [𝑨𝑵𝑺𝑾𝑬𝑹, 𝝀 𝒂𝒕 𝒘𝒉𝒊𝒄𝒉 𝒎(𝝀)𝒊𝒔 𝒍𝒂𝒓𝒈𝒆𝒔𝒕] [Note: This has to be the answer as there are no other critical points to choose from.] Hence, the maximum value of 𝑚(𝜆) 𝑚(𝜆)𝑚𝑎𝑥 = 𝑚(2) = −5(2)2 + 20(2) = 𝟐𝟎 [𝑨𝑵𝑺𝑾𝑬𝑹] c) Here, 𝑓(𝑥, 𝑦) = 𝑥 2 + 𝑦 2 [1] 𝑔(𝑥, 𝑦) = 2𝑥 + 4𝑦 [2] And Hence, using Lagrange multipliers, we have the constraints ∇𝑓 = 𝜆∇𝑔 And 2𝑥 + 4𝑦 = 20 [3] Thus, dividing the gradient operator to 2 components (x and y partial derivatives) ...
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