Expression "-3x^2+y4x" when x = "-4" and y = 2 A. -112 B. -80 C. 80 D. 272

A fence is to be built to enclose a rectangular area of 450 square feet. The fence along three sides is to be made of material t

Mila [183]

Answer:

The short side is _15___ft and the long side is ___30___ ft.

Step-by-step explanation:

As fence is built in a rectangular area, so we can consider

Let

x be the length of the rectangle

y be width of the rectangle

Given area of rectangle is = 450 ft²

Formula for area of rectangle = Length x width

450 ft² = xy

Solve for y

y = 450/x

now according to given condition

three sides of the fence costs $5 per foot and for the fourth side costs$15 per foot.

We have two condition either the fourth side be x or y

So condition 1: Three sides =(x,y,x) 4-th side = y.

So we can write as 5x,5y,5x and 15 y

Cost C = 5x +5y+5x+15y

= 10x+ 5y+15y

= 5(2x+y) +15y----------------equation 1

= 10x +20y

Adding value of y 450/x

= 10x + 20(450/x)

= 10x + 9000/x

For minimum cost, we can consider the cost to be 0

0 = 10x + 9000/x

Dividing and multiplying by -x/x

0 = -10 +9000/x²

10 = 9000/x²

10x² = 9000/

x²= 900

x = 30

so y = 450/x = 450/30= 15 ft

so adding the values of x and y in equation 1 we will have

cost C= 5(2x+y) +15y----------------equation 1

cost is = 5(2(30)+15) +15(15)

= $600 is the cost

X= 30 y =15

So condition 2: Three sides =(y,x,y) 4-th side = x.

So we can write as 5y,5x,5y and 15 x

Cost C = 5y +5x+5y+15x

= 5x+ 10y+15x

= 5(x+2y) +15x----------------equation 2

= 20x +10y

Adding value of y= 450/x

= 20x + 10(450/x)

= 20x + 4500/x

For minimum cost, we can consider the cost to be 0

0 = 20x + 4500/x

Dividing and multiplying by -x/x

0 = -20 +4500/x²

20 = 4500/x²

20x² = 4500

x²= 4500/20= 225

x = 15

so y = 450/x = 450/15= 30 ft

so adding the values of x and y in equation 2 we will have

cost C= = 5(x+2y) +15x----------------equation 2

cost is = 5(15+2(30) +15(15)

= $600 is the cost

y= 30 x=15

so from both conditions satisfy the cost and the two sides are known as length and width

so dimension will be 15 ft by 30 ft

7 0

3 years ago

Lagrange multipliers have a definite meaning in load balancing for electric network problems. Consider the generators that can o

Ivahew [28]

Answer:

The load balance $(x_1,x_2,x_3)=(545.5,272.7,181.8)$ Mw minimizes the total cost

Step-by-step explanation:

Optimizing With Lagrange Multipliers

When a multivariable function f is to be maximized or minimized, the Lagrange multipliers method is a pretty common and easy tool to apply when the restrictions are in the form of equalities.

Consider three generators that can output xi megawatts, with i ranging from 1 to 3. The set of unknown variables is x1, x2, x3.

The cost of each generator is given by the formula

$\displaystyle C_i=3x_i+\frac{i}{40}x_i^2$

It means the cost for each generator is expanded as

$\displaystyle C_1=3x_1+\frac{1}{40}x_1^2$

$\displaystyle C_2=3x_2+\frac{2}{40}x_2^2$

$\displaystyle C_3=3x_3+\frac{3}{40}x_3^2$

The total cost of production is

$\displaystyle C(x_1,x_2,x_3)=3x_1+\frac{1}{40}x_1^2+3x_2+\frac{2}{40}x_2^2+3x_3+\frac{3}{40}x_3^2$

Simplifying and rearranging, we have the objective function to minimize:

$\displaystyle C(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)$

The restriction can be modeled as a function g(x)=0:

$g: x_1+x_2+x_3=1000$

Or

$g(x_1,x_2,x_3)= x_1+x_2+x_3-1000$

We now construct the auxiliary function

$f(x_1,x_2,x_3)=C(x_1,x_2,x_3)-\lambda g(x_1,x_2,x_3)$

$\displaystyle f(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)-\lambda (x_1+x_2+x_3-1000)$

We find all the partial derivatives of f and equate them to 0

$\displaystyle f_{x1}=3+\frac{2}{40}x_1-\lambda=0$

$\displaystyle f_{x2}=3+\frac{4}{40}x_2-\lambda=0$

$\displaystyle f_{x3}=3+\frac{6}{40}x_3-\lambda=0$

$f_\lambda=x_1+x_2+x_3-1000=0$

Solving for \lambda in the three first equations, we have

$\displaystyle \lambda=3+\frac{2}{40}x_1$

$\displaystyle \lambda=3+\frac{4}{40}x_2$

$\displaystyle \lambda=3+\frac{6}{40}x_3$

Equating them, we find:

$x_1=3x_3$

$\displaystyle x_2=\frac{3}{2}x_3$

Replacing into the restriction (or the fourth derivative)

$x_1+x_2+x_3-1000=0$

$\displaystyle 3x_3+\frac{3}{2}x_3+x_3-1000=0$

$\displaystyle \frac{11}{2}x_3=1000$

$x_3=181.8\ MW$

And also

$x_1=545.5\ MW$

$x_2=272.7\ MW$

The load balance $(x_1,x_2,x_3)=(545.5,272.7,181.8)$ Mw minimizes the total cost

5 0

4 years ago

I need help pls and thank you

Aleks [24]

The fourth table is the correct answer

6 0

3 years ago

(3a) You spend $72 on tapes and CDs. Each tape costs $8 and each CD costs $12.

Makovka662 [10]

Answer:

3a. 8t + 12c = 72

3b. t + c = 7

Step-by-step explanation:

number of tapes is t

number of Cds is c

so 8t + 12c = 72

and t + c = 7

so t = 7 - c

replace t = 7 - c in to the first equation

8(7 - c) + 12c = 72

56 - 8c + 12c = 72

4c = 72 - 56 = 16

c = 4

if c = 4, then t = 7 - c = 7 - 4 = 3

6 0

2 years ago

What is the value of x

Alex

Answer:

IDK please add a picture/equation so I can understand ty

Step-by-step explanation:

3 0

3 years ago