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1 year ago

7

a famer has 120 feet of fencing with whcih to enclose two adjacent rectangular pens as shown. what dimeensions should be used th

at the enclosed area will be a maximum? what will the area be?

1 answer:

Nat2105 [25]1 year ago

6 0

The dimensions of the rectangular pen should be 15 by 20 feet and the maximum area is 1200 square feet.

Let the area be y .

Area = (base) × (height)

Base = 2x

Height = h

Let the area of the rectangular pens be y .

∴ y = 2xh

Perimeter of all the fencing = 4x+3h

∴ 4x+3h = 120

now we solve for h

3h = 120-4x

h = 40 - 4/3 x

Now we will substitute this value in the above first equation:

y = 2xh

or, y = 2x (40 - 4/3 x)

or, y = 80x - 8/3 x²

Now for the maximum area we have to find the first order differentiation of y

now,

dy /dx = 80 - 16/3 x

At dy/dx = 0 we get the value of x for which y is maximum.

80 - 16/3 x = 0

or, - 16/3 x = -80

or, x = 15 feet

Hence height = 40 - 4/3 x = 40 - 20 = 20feet

Maximum area = 2xh = 2×15×40 = 1200 square feet

The dimensions of the rectangular pen should be 15 by 20 feet and the maximum area is 1200 square feet.

Disclaimer : The missing figure for the question is attached below.

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A graphing calculator is recommended. A function is given. g(x) = x4 − 5x3 − 14x2 (a) Find all the local maximum and minimum val

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Answer:

The local maximum and minimum values are:

Local maximum

$g(0) = 0$

Local minima

$g(5.118) = -350.90$

$g(-1.368) = -9.90$

Step-by-step explanation:

Let be $g(x) = x^{4}-5\cdot x^{3}-14\cdot x^{2}$ . The determination of maxima and minima is done by using the First and Second Derivatives of the Function (First and Second Derivative Tests). First, the function can be rewritten algebraically as follows:

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Then, first and second derivatives of the function are, respectively:

First derivative

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$g'(x) = 2\cdot x^{3}-10\cdot x^{2}-28\cdot x +2\cdot x^{3}-5\cdot x^{2}$

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$g'(x) = x\cdot (4\cdot x^{2}-15\cdot x -28)$

Second derivative

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Now, let equalize the first derivative to solve and solve the resulting equation:

$x\cdot (4\cdot x^{2}-15\cdot x -28) = 0$

The second-order polynomial is now transform into a product of binomials with the help of factorization methods or by General Quadratic Formula. That is:

$x\cdot (x-5.118)\cdot (x+1.368) = 0$

The critical points are 0, 5.118 and -1.368.

Each critical point is evaluated at the second derivative expression:

x = 0

$g''(0) = 12\cdot (0)^{2}-30\cdot (0) -28$

$g''(0) = -28$

This value leads to a local maximum.

x = 5.118

$g''(5.118) = 12\cdot (5.118)^{2}-30\cdot (5.118) -28$

$g''(5.118) = 132.787$

This value leads to a local minimum.

x = -1.368

$g''(-1.368) = 12\cdot (-1.368)^{2}-30\cdot (-1.368) -28$

$g''(-1.368) = 35.497$

This value leads to a local minimum.

Therefore, the local maximum and minimum values are:

Local maximum

$g(0) = (0)^{4}-5\cdot (0)^{3}-14\cdot (0)^{2}$

$g(0) = 0$

Local minima

$g(5.118) = (5.118)^{4}-5\cdot (5.118)^{3}-14\cdot (5.118)^{2}$

$g(5.118) = -350.90$

$g(-1.368) = (-1.368)^{4}-5\cdot (-1.368)^{3}-14\cdot (-1.368)^{2}$

$g(-1.368) = -9.90$

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