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SSSSS [86.1K]
3 years ago
6

Use the expression 5(6 + 4x) to answer the following:

Mathematics
1 answer:
marissa [1.9K]3 years ago
3 0
The answer is a the answer is a
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Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d
andrey2020 [161]

Answer:

The maximum value is 1/27 and the minimum value is 0.

Step-by-step explanation:

Note that the given function is equal to (xyz)^2 then it means that it is positive i.e f(x,y,z)\geq 0.

Consider the function F(x,y,z,\lambda)=x^2y^2z^2-\lambda (x^2+y^2+z^2-1)

We want that the gradient of this function  is equal to zero. That is (the calculations in between are omitted)

\frac{\partial F}{\partial x} = 2x(y^2z^2 - \lambda)=0

\frac{\partial F}{\partial y} = 2y(x^2z^2 - \lambda)=0

\frac{\partial F}{\partial z} = 2z(x^2y^2 - \lambda)=0

\frac{\partial F}{\partial \lambda} = (x^2+y^2+z^2-1)=0

Note that the last equation is our restriction. The restriction guarantees us that at least one of the variables is non-zero. We've got 3 options, either 1, 2 or none of them are zero.

If any of them is zero, we have that the value of the original function is 0. We just need to check that there exists a value for lambda.

Suppose that x is zero. Then, from the second and third equation we have that

-2y\lambda = -2z\lambda. If lambda is not zero, then y =z. But, since -2y\lambda=0 and lambda is not zero, this implies that x=y=z=0 which is not possible. This proofs that if one of the variables is 0, then lambda is zero. So, having one or two variables equal to zero are feasible solutions for the problem.

Suppose that only x is zero, then we have the solution set y^2+z^2=1.

If both x,y are zero, then we have the solution set z^2=1. We can find the different solution sets by choosing the variables that are set to zero.

NOw, suppose that none of the variables are zero.

From the first and second equation we have that

\lambda = y^2z^2 = x^2z^2 which implies x^2=y^2

Also, from the first and third equation we have that

\lambda = y^2z^2 = x^2y^2 which implies x^2=z^2

So, in this case, replacing this in the restriction we have 3z^2=1, which gives as another solution set. On this set, we have x^2=y^2=z^2=\frac{1}{3}. Over this solution set, we have that the value of our function is \frac{1}{3^3}= \frac{1}{27}

4 0
3 years ago
Which is the lower rate?<br> 44 people in 4 theaters or 100 people in 10 theaters
madam [21]

Answer:

100 in 10 theaters. 100÷10=10 44÷4=11

3 0
3 years ago
20 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
dybincka [34]
X²+y²=r²   ⇒ r²=9  ⇒ r=3

radius = 3.
3 0
3 years ago
Read 2 more answers
Slope of a parallel line with the points (7,-5) and (-5,7)
VLD [36.1K]

Answer: Negative 1  The slope of parallel lines is the same.

The attachment shows two such lines, given coordinates labeled.

Step-by-step explanation:

Find the slope of the line passing through the given points.

 rise/run  

Rise is the difference in y-values  7-(-5)  = 12

Run is the difference between x-values  -5 - 7  = - 12

The Slope is 12/-12   simplify:

slope = -1

5 0
3 years ago
Evaluate \dfrac j4 4 j ? start fraction, j, divided by, 4, end fraction when j=12j=12j, equals, 12.
cricket20 [7]

We need to evaluate the value of fraction \frac{j}{4}.

We also given j=12.

In order to evaluate the value of fraction \frac{j}{4}, we need to plug j=12.

Plugging j=12, we get

\frac{j}{4}= \frac{12}{4}

We have 12 in numerator and 4 in denominator.

We always divide top number by bottom number.

So, we need to divide 12 by 4.

On dividing 12 by 4 we get 3.

<h3>Therefore, \frac{j}{4} = 3.</h3>
3 0
3 years ago
Read 2 more answers
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