Total of 49 1/4 + 3 3/8?

The temperature function (in degrees Fahrenheit) in a three dimensional space is given by T(x, y, z) = 3x + 6y - 6z + 1. A bee i

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).

6 0

3 years ago

Find the values of m and b that make the following function differentiable.

dimaraw [331]

Now, bearing in mind that, for a graph, to be "differentiable", means a "smooth transition", not an abrupt edge or a "cusp", the first subfunction of the piece-wise is just a quadratic, the second subfunction is a linear, well, so the linear needs to hit the quadratic "smoothly", not make an abrupt edge, for that, the linear must continue the quadratic from 2 onwards, since 2 is the range edge.

now, what is the piece-wise when x = 2? well, f(2) = x² or (2)² or 4

that means, the linear needs to hit it at that point, to make it so, let's make it the slope of the linear, the same as the quadratic's.

what's the quadratic's slope? well, simple enough

$\bf \left. \cfrac{dy}{dx}=2x \right|_{x=2}\implies 4\impliedby \textit{this means }
\begin{array}{llll}
mx+b\\
\uparrow \\
4
\end{array}$

so we have y = 4x + b

now, recall, for the quadratic, when x =2, y = 4, so for the linear, when x = 2, y = 4 as well, this means

$\bf 4 = 4(2) + b\implies 4=8+b\implies 4-8=b\implies \boxed{-4=b}
\\\\\\
mx+b\implies 4x-4\\\\
-------------------------------\\\\
f(x)=
\begin{cases}
x^2&x\le 2\\
4x-4&x\ \textgreater \ 2
\end{cases}$

check the picture below.

4 0

3 years ago

What is the area of the following parallelogram in square centimeters?

Scrat [10]

Answer:

3.6 cm²

Step-by-step explanation:

18 mm = 1.8 cm

A = 2cm×1.8cm = 3.6 cm²

7 0

3 years ago

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Two airplanes leave an airport at the same time, the first headed due north and the second at a bearing of N42^ E . At 2:00PM, t

Leno4ka [110]

Answer:

The two airplanes are about 330miles apart.

Step-by-step explanation:

The diagram interpreting the question has been attached to this response.

As shown in the diagram,

i. the airplanes leave at point C.

ii. at 2.00pm the first and second airplanes are at points A and B respectively, where they are 312miles and 487miles away from the starting point C in directions due north and N42E from the point C.

iii. the points A, B and C form a triangle with sides a, b and c.

To solve for the value of c which is the distance between the two planes at 2.00pm, the cosine rule is used.

c² = a² + b² - 2abcosC --------------(i)

where;

b = 312miles

a = 487miles

C = 42°

Substitute these values into equation (i) and solve as follows;

c² = (487)² + (312)² - 2(312)(487)cos(42)

c² = (237169) + (97344) - 303888cos(42)

c² = (237169) + (97344) - 303888(0.7431)

c² = 334513 - 225819.1728

c² = 108693.8272

<em>Take the square root of both sides</em>

√c² = √108693.8272

c = 329.69

c ≅ 330miles

Therefore, the two airplanes are far apart by 330miles

3 0

3 years ago

Please and thank you

algol [13]

Answer:

the last one

Step-by-step explanation:

6 0

3 years ago

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