MATHEMATICS MIDDLE SCHOOL

A school wishes to enclose its rectangular playground using 480 meters of fencing.Suppose that a side length (in meters) of the playground is , as shown below.

(a) Find a function that gives the area A(x) of the playground (in square meters) in terms of x.

(b) What side length x gives the maximum area that the playground can have?

(c) What is the maximum area that the playground can have?

Answers

Answer 1

Answer:

Answer:

Part a) $A(x)=(-x^2+240x)\ m^2$

Part b) The side length x that give the maximum area is 120 meters

Part c) The maximum area is 14,400 square meters

Step-by-step explanation:

The picture of the question in the attached figure

Part a) Find a function that gives the area A(x) of the playground (in square meters) in terms of x

we know that

The perimeter of the rectangular playground is given by

$P=2(L+W)$

we have

$P=480\ m\nL=x\ m$

substitute

$480=2(x+W)$

solve for W

$240=x+W\nW=(240-x)\ m$

Find the area of the rectangular playground

The area is given by

$A=LW$

we have

$L=x\ m\nW=(240-x)\ m$

substitute

$A=x(240-x)\nA=-x^2+240x$

Convert to function notation

$A(x)=(-x^2+240x)\ m^2$

Part b) What side length x gives the maximum area that the playground can have?

we have

$A(x)=-x^2+240x$

This function represent a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

The x-coordinate of the vertex represent the length that give the maximum area that the playground can have

Convert the quadratic equation into vertex form

$A(x)=-x^2+240x$

Factor -1

$A(x)=-(x^2-240x)$

Complete the square

$A(x)=-(x^2-240x+120^2)+120^2$

$A(x)=-(x^2-240x+14,400)+14,400$

$A(x)=-(x-120)^2+14,400$

The vertex is the point (120,14,400)

therefore

The side length x that give the maximum area is 120 meters

Part c) What is the maximum area that the playground can have?

we know that

The y-coordinate of the vertex represent the maximum area

The vertex is the point (120,14,400) -----> see part b)

therefore

The maximum area is 14,400 square meters

Verify

$x=120\ m$

$W=(240-120)=120\ m$

The playground is a square

$A=120^2=14,400\ m^2$

Answer 2

Answer:

Final answer:

The width of the playground is 120 meters, the side length that gives the maximum area is 120 meters, and the maximum area the playground can have is 14400 square meters.

Explanation:

(a) Let's assume the width of the rectangle is x meters. Since the playground is rectangular and has two equal sides, the length will also be x meters. The perimeter of the rectangle, which is also the amount of fencing needed, is given as 480 meters. This can be expressed as: 2(length + width) = 480. Using this equation, we can solve for the width: 2(x + x) = 480 ⇒ 4x = 480 ⇒ x = 480/4 = 120. Therefore, the width of the playground is 120 meters.

(b) To find the side length that gives the maximum area, we can use calculus. The area function is A(x) = x * x = x^2. To find the maximum of this function, we can take the derivative and set it equal to zero: dA/dx = 2x = 0 ⇒ x = 0. So, x = 0 is a critical point, but since we are dealing with a physical situation where the length cannot be zero, we disregard this critical point. Thus, x = 120 is the value that gives the maximum area.

(c) Now that we know the side length, we can calculate the maximum area. Plugging in x = 120 into the area function, we find: A(120) = 120 * 120 = 14400 square meters. Therefore, the maximum area the playground can have is 14400 square meters.

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