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Stella [2.4K]
3 years ago
14

1.The area of the KM 89 Farm in San Vicente is (x²+7x +12) sq meters. If its width is (x + 3) meters, what is the length of the

farm?
2. A box has a volume of (x³ + 12x²+ 47x + 60) cu. cm. If it is (x + 3) cm high, what is the product its width and length?

3. The base of a parallelogram is(f-2) cm. If its area is (f³-8) cm², find an expression for the height of the parallelogram.4. Any wants to paint the wall of his room for renovation. If the area of the wall is (3x³- 4x² + 2x + 4) m² and a paint can cover (x²- 2x + 2) m², how many cans of paint will he use to paint the whole wall?

5. A rectangular garden in a backyard has an area of (x² + 5x - 6) square meters. Its width is (x + 2) meters. What is the length of the garden?

I need the solution with the answer thank youuu
Mathematics
1 answer:
OLga [1]3 years ago
4 0

Step-by-step explanation:

1. A = lw

  • l = A/w
  • l = (x²+7x +12) / (x + 3) =
  • (x + 4)(x + 3)/(x + 3) =
  • x + 4 m

2. V = lwh

  • lw = V/h
  • lw = (x³ + 12x²+ 47x + 60) / (x + 3) =
  • (x³+ 3x² + 9x² + 27x + 20 x + 60) /(x + 3) =
  • (x + 3)(x² + 9x + 20) / (x + 3) =
  • x² + 9x + 20 cm²

3. A = bh

  • h = A/b
  • h = (f³-8) / (f-2) =
  • (f - 2)(f² + 2f + 4)/(f-2) =
  • f² + 2f + 4 cm

4. A = p*c

  • c = A/p
  • c =  (3x³- 4x² + 2x + 4) / (x²- 2x + 2) =
  • (3x³ - 6x² + 6x + 2x² - 4x + 4) /(x² - 2x + 2) =
  • (x² - 2x + 2) (3x + 2) / (x² - 2x + 2) =
  • 3x + 2 cans

5. A = lw

  • l = A/w
  • l = (x² + 5x - 6)/(x + 2) =
  • (x - 1)(x + 6)/(x + 2) m

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For given f(x, y) the extremum: (12, 24) which is the minimum.

For given question,

We have been given a function f(x) = 4x² + 2y² under the constraint 3x+3y= 108

We use the constraint to build the constraint function,

g(x, y) = 3x + 3y

We then take all the partial derivatives which will be needed for the Lagrange multiplier equations:

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Setting up the Lagrange multiplier equations:

f_x=\lambda g_x

⇒ 8x = 3λ                                        .....................(1)

f_y=\lambda g_y

⇒ 4y = 3λ                                         ......................(2)

constraint: 3x + 3y = 108                .......................(3)

Taking (1) / (2), (assuming λ ≠ 0)

⇒ 8x/4y = 1

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Substitute this value of y in equation (3),

⇒ 3x + 3y = 108

⇒ 3x + 3(2x) = 108

⇒ 3x + 6x = 108

⇒ 9x = 108

⇒ x = 12

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So, the saddle point (critical point) is (12, 24)

Now we find the value of f(12, 24)

⇒ f(12, 24) = 4(12)² + 2(24)²

⇒ f(12, 24) = 576 + 1152

⇒ f(12, 24) = 1728                             ................(1)

Consider point (18,18)

At this point the value of function f(x, y) is,

⇒ f(18, 18) = 4(18)² + 2(18)²

⇒ f(18, 18) = 1296 + 648

⇒ f(18, 18) = 1944                            ..............(2)

From (1) and (2),

1728 < 1944

This means, given extremum (12, 24) is minimum.

Therefore, for given f(x, y) the extremum: (12, 24) which is the minimum.

Learn more about the extremum here:

brainly.com/question/17227640

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