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Svet_ta [14]
2 years ago
5

Use Lagrange multipliers to find the maximum and minimum values of f(x,y)=x+4y subject to the constraint x2+y2=9, if such values

exist. Round your answers to three decimal places. If there is no global maximum or global minimum, enter NA in the appropriate answer area. Maximum =
Mathematics
1 answer:
Vesnalui [34]2 years ago
4 0

The Lagrangian is

L(x,y,\lambda)=x+4y+\lambda(x^2+y^2-9)

with critical points where the partial derivatives vanish.

L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}

L_y=4+2\lambda y=0\implies y=-\dfrac2\lambda

L_\lambda=x^2+y^2-9=0

Substitute x,y into the last equation and solve for \lambda:

\left(-\dfrac1{2\lambda}\right)^2+\left(-\dfrac2\lambda\right)^2=9\implies\lambda=\pm\dfrac{\sqrt{17}}6

Then we get two critical points,

(x,y)=\left(-\dfrac3{\sqrt{17}},-\dfrac{12}{\sqrt{17}}\right)\text{ and }(x,y)=\left(\dfrac3{\sqrt{17}},\dfrac{12}{\sqrt{17}}\right)

We get an absolute maximum of 3\sqrt{17}\approx12.369 at the second point, and an absolute minimum of -3\sqrt{17}\approx-12.369 at the first point.

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1-tan^2(x)/sec^2 = cos(2x)
LekaFEV [45]

By applying the formula of trigonometric function the right hand side will

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Taking right hand side first which is cos2x.

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Now we will solve the left hand side of the equation give which is

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1-tan^{2}x/1+tan^{2}x=1-tan^{2}x/1+tan^{2}x.

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1 year ago
Evaluate and reduce:<br> 12 + (12)<br> Enter the number that belongs in the<br> green box.<br> [?]
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12 + (12)

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I think of a number , multiply it by 3, add 4 and square the result
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4 0
2 years ago
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Please use statement and reason, not sin cos or tan. In rectangle ABCD the diagonals intersect each other at point O and ABD = 3
zvonat [6]

Answer:

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Step-by-step explanation:

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<h3>Given</h3>
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<h3>Solution</h3>

<u>As per properties mentioned above we have:</u>

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Side opposite to 30 is half of the length of the hypotenuse

<u>AD is opposite to ∠B and BD is the hypotenuse, then:</u>

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and

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