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Paha777 [63]
2 years ago
10

3) Puan wishes to build a garden at the bottom of his property. He wants to split in

Mathematics
1 answer:
frozen [14]2 years ago
3 0

The area of the garden is the product of its dimensions

  • The dimension that maximizes the area is 20 by 60
  • The maximum area is 1200 square feet

From the figure (see attachment), the perimeter is:

\mathbf{P  = 3x + y}

The perimeter is given as: 120.

So, we have:

\mathbf{3x + y = 120}

Make y the subject

\mathbf{y = 120 - 3x}

The area is then calculated as:

\mathbf{A =xy}

Substitute \mathbf{y = 120 - 3x}

\mathbf{A =x(120 - 3x)}

\mathbf{A =120x- 3x^2}

Differentiate

\mathbf{A' =120- 6x}

Set to 0

\mathbf{120- 6x = 0}

Rewrite as:

\mathbf{6x = 120}

Divide through by 6

\mathbf{x = 20}

Recall that: \mathbf{y = 120 - 3x}

So, we have:

\mathbf{y = 120 - 3(20)}

\mathbf{y = 120 - 60}

\mathbf{y =60}

The dimension that maximizes the area is 20 by 60

The area is calculated as:

\mathbf{A =xy}

So, we have:

\mathbf{A = 20 \times60}

\mathbf{A = 1200}

Hence, the maximum area is 1200 square feet

Read more about maximum areas at:

brainly.com/question/11906003

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\left[\begin{array}{ccccc}1&P&0&0&0\\1-P&1&P&0&0\\0&1-P&1&P&0\\0&0&1-P&1&P\\0&0&0&1-P&1\end{array}\right]

Calculating the limiting probabilities:

π0=π0+Pπ1                 eq(1)

π1=(1-P)π0+π1+Pπ2     eq(2)

π2=(1-P)π1+π2+Pπ3    eq(3)

π3=(1-P)π2+π3+Pπ4    eq(4)

π4=(1-P)π3+π4             eq(5)

π0+π1+π2+π3+π4=1

π0-π0-Pπ1=0

→π1 = 0

substituting value of π1  in eq(2)

(1-P)π0+Pπ2=0

from

π2=(1-P)π1+π2+Pπ3  

we get

(1-P)π1+Pπ3 = 0

from

π3=(1-P)π2+π3+Pπ4

we get

(1-P)π2+Pπ4 =0

from π4=(1-P)π3+π4  

→π3=0

substituting values of π1 and π3 in eq(3)

→π2=0

Now

π0+π1+π2+π3+π4=0

π0+π4=1

π0=0.5

π4=0.5

So limiting probabilities are {0.5,0,0,0,0.5}

4 0
3 years ago
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