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11 days ago

A rectangular schoolyard is to be fenced in using the wall of the school for one side and 150

Mathematics

1 answer:

lawyer [4K]11 days ago

5 0

Answer:

Part 1) The equation is $A(x)=150x-2x^2$

Part 2) When x=40 m, the area of the schoolyard is A=2,800 m^2

Part 3) The valid domain consists of all real numbers exceeding zero and below 75 meters

Step-by-step explanation:

Part 1) Formulate an expression for A(x)

Let

x -----> the length of the rectangular school yard

y ---> the width of the rectangular school yard

It is known that

The perimeter for the fencing (taking the school wall as one side) is

$P=2x+y$

$P=150\ m$

thus

$150=2x+y$

$y=150-2x$ -----> this is equation A

The area of the rectangular school yard is

$A=xy$ ----> this is equation B

Substituting equation A into equation B yields

$A=x(150-2x)$

$A=150x-2x^2$

Change to function notation

$A(x)=150x-2x^2$

Part 2) What is the area when x=40?

With x equal to 40 m

substitute the value into the expression from Part 1 to determine A

$A(40)=150(40)-2(40^2)$

$A(40)=2,800\ m^2$

Part 3) What would be a suitable domain for A(x) in this scenario?

We understand that

A signifies the area of the rectangular school yard

x characterizes the length of the rectangular school yard

It follows that

$A(x)=150x-2x^2$

This forms a vertical parabola opening downwards

The vertex indicates a maximum point

The x-coordinate of the vertex corresponds to the length that maximizes the area

The y-coordinate of the vertex denotes the maximum area

The vertex corresponds to (37.5, 2812.5)

Refer to the accompanying figure

Consequently,

The peak area achieved is 2,812.5 m^2

The x-intercepts are located at x=0 m and x=75 m

The domain for A is the range -----> (0, 75)

All real numbers greater than zero and less than 75 meters

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Response:

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Step-by-step breakdown:

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Substituting the known values,

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Now we'll input the values.

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We substitute values here as well.

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