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allochka39001 [22]
1 year ago
12

guse lagrange multipliers to find the maximum or minimum values of the function subject to the given constraint. (if an answer d

oes not exist, enter dne.) f(x, y)
Mathematics
1 answer:
Novosadov [1.4K]1 year ago
3 0

Therefore the maximum value of function f(x,y,z)=x^{2} y^{2} z^{2} =1/27

And the minimum value is 0

<h3>What is function?</h3>

function, in mathematics, an expression, rule, or law that defines a relationship between one variable (the independent variable) and another variable (the dependent variable) (the dependent variable) (the dependent variable) (the dependent variable). Mathematics uses functions frequently, and functions are essential for specifying physical relationships in the sciences.

Here,

The function is given as:

f(x,y,z)=x^{2} y^{2} z^{2}

x^{2} +y^{2}+ z^{2}=1

=>x^{2} +y^{2}+ z^{2}-1=0

Using Lagrange multiplies, we have:

L(x,y,z,λ)=f(x,y,z) +λ(0)

Substitute f(x,y,z)=x^{2} y^{2} z^{2}  and x^{2} +y^{2}+ z^{2}-1=0

Differentiate

L(x)=2xy^{2} z^{2}+2λx

L(y)=2yx^{2} z^{2}+2λy

L(z)=2zx^{2} y^{2}+2λz

L(λ)=x^{2} +y^{2}+ z^{2}-1

Equating to 0

2xy^{2} z^{2}+2λx =0

2yx^{2} z^{2}+2λy = 0

2zx^{2} y^{2}+2λz = 0

x^{2} +y^{2}+ z^{2}-1 = 0

Factorize the above expressions

2xy^{2} z^{2}+2λx =0

2x(y^{2} z^{2}+λ)=0

2x=0 and (y^{2} z^{2}+λ)=0

x=0 and  y^{2} z^{2}= -λ

2yx^{2} z^{2}+2λy = 0

2y(x^{2} z^{2}+λ)=0

2y=0 and (x^{2} z^{2}+λ)=0

y=0 and  x^{2} z^{2}= -λ

2zx^{2} y^{2}+2λz = 0

2z(y^{2} x^{2}+λ)=0

2z=0 and (x^{2} y^{2}+λ)=0

z=0 and  x^{2} y^{2}= -λ

So we have ,

x=0 and  y^{2} z^{2}= -λ

y=0 and  x^{2} z^{2}= -λ

z=0 and  x^{2} y^{2}= -λ

The above expression becomes

x=y=z=0

This means that,

x^{2} +y^{2}+ z^{2}=1

x^{2} +x^{2}+ x^{2} =1 \\3x^{2 } =1

x= ±1/\sqrt{3}

So,

y= ±1/\sqrt{3}

z= ±1/\sqrt{3}

The critic points are

x=y=z=±1/\sqrt{3}

x=y=z=0

Therefore the maximum value of function f(x,y,z)=x^{2} y^{2} z^{2} =1/27

And the minimum value is 0

To know  more about function , visit

brainly.com/question/12426369

#SPJ4

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A trader sold a radio set for 72.00 making a profit of 8% .Find , correct to the nearest ghana cedis ,the cost price for the rad
VMariaS [17]

Given:

The selling price of a ratio set = 72.00

Profit percent = 8%

To find:

The cost price for the radio set.

Solution:

Let x be the cost price for the radio set.

Radio set is sold at 8% profit.

\text{Selling price}=x+\dfrac{8}{100}x

\text{Selling price}=x+0.08x

\text{Selling price}=1.08x

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72=1.08x

\dfrac{72}{1.08}=x

x=66.666...

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Therefore, the cost price for the radio set is 67.

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Is there a commutative property of subtraction that states a-b=b-a? investigate and make a conclusion that you justify.
Rom4ik [11]

No is your answer

Assuming that b ≠ a, the answers will not be the same.

For example, (remembering that b ≠ a) let us assume that b = 10, a = 5

10 - 5 = 5

5 - 10 = -5

5 ≠ -5

So the commutative property of subtraction does not work unless in certain cases, in which a = b.

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A rectangle has perimeter 28cm. Its area is 42sq.cm. Determine the dimensions of the rectangle. Include a diagram in your soluti
yawa3891 [41]

The dimensions of the rectangle are: b=9.65 cm and h=4.35 cm or b=4.35 cm and h= 9.65 cm.

<h3>Quadrilaterals</h3>

There are different types of quadrilaterals, for example, square, rectangle, rhombus, trapezoid, and parallelogram.  Each type is defined accordingly to its length of sides and angles. For example, in a rectangle,  the opposite sides are equal and parallel and their interior angles are equal to 90°.

The area of a rectangle is equal to base x height (A=bh). The perimeter of a geometric figure is the sum of its sides. Thus, for a rectangle, the perimeter is 2b+2h.

<h3>System of Linear Equations</h3>

System of linear equations is the given term math for two or more equations with the same variables. The solution of these equations represents the point at which the lines intersect.

The question gives:

Perimeter=28cm

Area= 42 cm²

If the perimeter is the sum of all sides of the rectangle, you have:

Perimeter= 2b+2h=28

If the area of the rectangle is 42cm², you have:

Area=bh=42 cm²

You can write a system of linear equations.

2b+2h=28 (1)

bh=42 (2)

From equation 2, you have  b=\frac{42}{h}. Then, you can replace it in equation 1.

2*\frac{42}{h}+2h=28 \\ \\ 84+2h^2=28h\\ \\ 2h^2-28h+84=0 \; dividing\; by\; 2\\ \\ h^2-14h+42=0

Now, you should solve the quadratic equation.

h_{1,\:2}=\frac{-\left(-14\right)\pm \sqrt{\left(-14\right)^2-4\cdot \:1\cdot \:42}}{2\cdot \:1}

h_{1,\:2}=\frac{-\left(-14\right)\pm \:2\sqrt{7}}{2\cdot \:1}

h_1=\frac{-\left(-14\right)+2\sqrt{7}}{2\cdot \:1}=7+\sqrt{7}=9.65 cm

h_2=\frac{-\left(-14\right)-2\sqrt{7}}{2\cdot \:1}=7-\sqrt{7}=4.35cm

If h1=9.65 cm, then b_1=\frac{42}{9.65} =4.35 cm, and if  h2=4.35 cm, then b_1=\frac{42}{4.35} =9.65 cm.

Read more about the quadratic function here:

brainly.com/question/1497716

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