ACT Math - Geometry and Optimization

Question

A rectangular playground is being built next to a school building, which will form one of the longer sides of the playground. If the school has 80 meters of fencing available to enclose the remaining three sides of the playground, what are the dimensions that will maximize the area of the playground? What is the maximum area? Infinite Craft

Answer 1

Answer 2

In order to answer this question you want to start by defining the variables. So if you let x represent the shorter side of the rectangle and y represent the longer side that runs parallel to the school building, then the perimeter is represented by the equation

2x + y = 80

Our goal then is to optimize the area, which would be

A(x) = x * (80 – 2x) = 80x – 2x^2

Because A(x) = -2x^2 + 80x is a parabola, the optimal value will be at the vertex, so you can take

x = -b/2a = -80 / 2(-2) = -80 / -4 = 20 in order to solve.

Then we can plug that back into the other expression to solve for the longer side!

2(20) + y = 80

40 + y = 80

y = 40

Meaning that the optimal side lengths would be 20 x 40 and then enclosed area would be 40 * 20 = 800 square meters.

I hope that helps :)

Answer 3

Let x be the width of a rectangle. Then its length is 80-2x. We are asked to maximize the area A(x)=x(80-2x)= -2(-40x+x^2)=-2(x^2-40x+400-400)=-2(x^2-40x+400)+800=-2(x-20)^2+800. The maximum is achived when x=20. Therefore, the width and length of the playground are 20 and 40, respectively.

Alternatively, we can view this geometrically. Consider the reflection of the rectangle with respect to the side that is the side of the school building. Then the two rectangles together form a larger one with the same length but doubled width 2x. Its perimeter is twice of the length of the total fence, which is 80x2=160. Now the area of the original playground is maximized exactly as the area of this "doubled" rectangle is maximized, which is achieved when it is a square. The side length of the square is thus 160/4=40, and the width of the original playground is half of the side length, 20.

Using Calculus (derivatives), we can solve it by taking the derivative of the are function A(x)=80x-2x^2. A'(x)=80-4x. When A(x) is maximized, A'(x)=0. Thus x=20, and 80-2x=40. The second derivative A''(x)=-4 implies that the function is concave and consequently A(x) has a maximum at x=20.

Answer 4

Where are the answer choices? ACT Math questions have answer choices.

Call the building length y. The 80 feet of fence have to make the two sides (call those x) and the side opposite the building, which is also y.

y + 2x = 80

Area =xy

From the first equation: y = 80-2x

Substitute that into second equation x(80-2x) = area = 80x - 2x^2

The max will be the vertex. ax^2 +bx + c the vertex is -b/2a

In this case, a = -2, and b = 80. Vertex = -80/2(-2) = -80/-4 = 20

y = 80(20) = 2(20)^2 = 1600 - 800 = 800.

Not sure which answer choice because those weren't provided.

Answer 5

Let y = length of the side parallel to the building

Let x = length of each of the other two sides

Then, 2x + y = 80

So, y = 80 - 2x

Area = xy = x(80 - 2x)

Graph is a parabola opening downward

x-intercepts: x = 0 and 80 - 2x = 0. So, x = 0, 40.

By the symmetry of the parabola, maximum area occurs when x is halfway between the x-intercepts.

Maximum area when x = 20, and y = 80 - 2x = 40.

Maximum area = xy = (20m)(40m) = 800 m²