F(x)= 25 + 4x How would you solve this please show your work. ASAP

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

Show how to to find the product 0.19 times 10 to the power of 3

rosijanka [135]

Answer:

190

Step-by-step explanation:

10x10x10=1000

then multiply 1000 by 0.19

190

3 0

3 years ago

PART A: A landmark on the first map is a triangle with side lengths of 3 cm, 4 cm, and 5 cm. What are the side lengths of the tr

Greeley [361]

Complete Question:

Johnny printed two maps of a walking trail near his home. The length of the walking trail on the first map is 8 cm.

(a) Choose a length between 5 cm and 15 cm for the walking trail on the second map: ________cm.

(b) Determine the scale factor from the first map to the second map.

(c) A landmark on the first map is a triangle with side lengths of 3 cm, 4 cm, and 5 cm. What are the side lengths of the triangle landmark on the second map?

(d) Draw one of the triangles from part C. Label the side lengths and vertices accordingly. You may use this drawing tool or draw your triangle on paper:

Answer:

(a) Length = 4cm

(b) Scale factor = 0.5

(c) 3cm, 4cm and 5cm are represented by 1.5cm, 2cm and 2.5cm respectively, on the second scale

(d) See attachment for triangle

Step-by-step explanation:

(a) Choose a length between 5 cm and 15 cm

$Length =4cm$ <em>--- This is solely up to you (you can make use of any length between 5 cm and 15 cm)</em>

(b) The scale factor

The scale factor (k) is calculated as:

$k = \frac{New\ Length}{Old\ Length}$

$k = \frac{4cm}{8cm}\\$

$k = 0.5$

(c) What are side lengths of 3 cm, 4 cm, and 5 cm on the second landmark

Using:

$k = \frac{New\ Length}{Old\ Length}$

The old lengths, in this case are: 3cm, 4cm and 5cm

Make New length the subject

$New\ Length = k * Old\ Length$

When $Length = 3cm$

$New\ Length = 0.5 * 3cm = 1.5cm$

When $Length = 4cm$

$New\ Length = 0.5 * 4cm = 2cm$

When $Length = 5cm$

$New\ Length = 0.5 * 5cm = 2.5cm$

So: 3cm, 4cm and 5cm are represented by 1.5cm, 2cm and 2.5cm respectively, on the second scale

(d) See attachment for triangle

7 0

3 years ago

Please help! How would you find the relative frequency of being female and attending an action movie?

denpristay [2]

Is this a STARR test?
A. Divide 99 by 250

6 0

3 years ago

Read 2 more answers

This question is from the tangent ratio section in our chapter and im not sure how to solve for this

VladimirAG [237]

The large triangle is an isosceles since both angles at the base each equal 42°.
In an isosceles triangle the altitude z is at the same time median , then it bisects the opposite side in the middle . So w = 120/2 = 60

Now let's calculate z:
tan 42° = (opposite side) / (adjacent side) = z/60
tan 42° = 0.9,
0.9 = z/60 and z = 54

6 0

3 years ago