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stellarik [79]
4 years ago
6

X^4-x^3-15x^2+9x+54=0

Mathematics
1 answer:
vekshin14 years ago
6 0

First write -x^3 as a difference

Your equation should look like this:

X^4➕2x^3➖3x^3➖15x^2➕9x➕54=0

Then write -15x^2 as a difference and write 9x as a sum

X^4➕2x^3➖33x^3➖6x^2➖9x^2➖18x➕27x➕54=0

Then factor out x^3 and -3x^2 along with -9x and 27

Then you factor out x➕2

(X➕2)✖️(x^3➖3x^2➖9x➕27)=0

Now factor out x^2 and -9

(x➕2)✖️(x^2✖️(x-3)-9(x-3)=0

Now factor out x-3 and split each into possible possibilities

x➕2=0

x➖3=0

x^2➖9=0

Then solve for x

You are then left with:

x1=-3, x2=-2, x3=3

Hope this helps! :3

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Answer:

The dimensions of the garden that minimize the cost is 9.018 feet(length) and 13.528 feet(width)

Step-by-step explanation:

Let the length of garden be x

Let the breadth of garden be y

Area of Rectangular garden = Length \times Breadth = xy

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So, xy=122 ---A

A landscape architect wished to enclose a rectangular garden on one side by a brick wall costing $20/ft

So, cost of brick along length x = 20 x

On the other three sides by a metal fence costing $10/ft.

So, Other three side s = x+2y

So, cost of brick along the other three sides= 10(x+2y)

So, Total cost = 20x+10(x+2y)=20x+10x+20y=30x+20y

Total cost = 30x+20y

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Total cost = 30x+20(\frac{122}{x})

Total cost = \frac{2440}{x}+30x

Now take the derivative to minimize the cost

f(x)=\frac{2440}{x}+30x

f'(x)=-\frac{2440}{x^2}+30

Equate it equal to 0

0=-\frac{2440}{x^2}+30

\frac{2440}{x^2}=30

\sqrt{\frac{2440}{30}}=x

9.018 =x

Now check whether it is minimum or not

take second derivative

f'(x)=-\frac{2440}{x^2}+30

f''(x)=-(-2)\frac{2440}{x^3}

Substitute the value of x

f''(x)=-(-2)\frac{2440}{(9.018)^3}

f''(x)=6.6540

Since it is positive ,So the x is minimum

Now find y

Substitute the value of x in A

(9.018)y=122

y=\frac{122}{9.018}

y=13.528

Hence the dimensions of the garden that minimize the cost is 9.018 feet(length) and 13.528 feet(width)

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