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melamori03 [73]
3 years ago
9

PLEASE HELPPPPPPPPPPPPPPP!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Vladimir79 [104]3 years ago
5 0
Your answer: 4,744,473,263,806,742
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madam [21]

You're looking for the extreme values of T(x,y,z)=3x+6y-6z+1 subject to x^2+y^2+z^2=9. The Lagrangian is

L(x,y,z,\lambda)=3x+6y-6z+1+\lambda(x^2+y^2+z^2-9)

with critical wherever the partial derivatives vanish:

L_x=3+2\lambda x=0\implies x=-\dfrac3{2\lambda}

L_y=6+2\lambda y=0\implies y=-\dfrac3\lambda

L_z=-6+2\lambda z=0\implies z=\dfrac3\lambda

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

Substituting the first three solutions into the last equation gives

\dfrac9{4\lambda^2}+\dfrac9{\lambda^2}+\dfrac9{\lambda^2}=9

\implies\lambda=\pm\dfrac32

\implies x=1,y=2,z=-2\text{ or }x=-1,y=-2,z=2

At these points, we have

T(1,2,-2)=28

T(-1,-2,2)=-26

so the highest temperature the bee can experience is 28º F at the point (1, 2, -2), and the lowest is -26º F at the point (-1, -2, 2).

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Step-by-step explanation:

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Solve for x<br> 510<br> (2x + 3)<br> 65%<br> 60°<br> 64°<br> 43<br> (3x - 1)º
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Given:
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X^2 - 4x - 77 = 0 Factor and check
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X² - 4x - 77 = 0

( x + 7) ( x - 11)

x + 7 = 0

x - 11 = 0

x = -7
x = 11
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3 years ago
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