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+1
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Find the minimum value of \(4 \cos \theta + \frac{49}{\cos \theta}\)

For \(0 \le \theta < \pi/2\)

Thanks!

 Dec 15, 2019
 #1
avatar
0

By the AM-GM inequality, \(4 \cos \theta + \frac{49}{\cos \theta} \ge 2 \sqrt{4 \cos \theta \cdot \frac{49}{\cos \theta}} = 28\)

 

The minimum value is 28.

 Dec 15, 2019
 #3
avatar+97 
+2

You can use the Am-GM, but what value makes 28 be the minimum

goreisthebest  Dec 15, 2019
 #2
avatar+129849 
+1

y = 4cosθ  +  49 / cosθ

 

y = 4cosθ  +  49secθ      take the derivative

 

y'  = -4sinθ  + 49secθtanθ

 

y'= -4sinθ  +  49sinθ /cos^2θ     set this to 0

 

-4sinθ  + 49sinθ /cos^θ  =  0

 

49sinθ / cos^2θ  =   4sinθ

 

49sinθ = 4sinθcos^2θ

 

49sinθ - 4sinθcos^2  =  0

 

sinθ  (49 - 4cos^2θ)  =  0

 

sinθ (7 - 2cosθ)(7 + 2cosθ)  =  0

 

So  either

 

sinθ  = 0            7  - 2cosθ =  0                                 7 + 2cosθ = 0

θ = 0                  7  = 2cosθ                                          2cosθ  = -7

                           cosθ = 7/2  ( impossible)                     cosθ = -7/2   (impossible)

 

So....the minimum  occurs when θ  =  0....

 

And the minimum  =   4cos(0)  + 49/cos(0)  =  4   + 49/1  =   53

 

As this graph shows  : https://www.desmos.com/calculator/42hzhstsx6

 

 

 

cool cool cool

 Dec 15, 2019

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