# Stuck on this maths question

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#1
Stephanie is making a wire sculpture. She bends a piece of thin wire of length 48cm into a Pentagon ABCDE

Angle AED= Angle CDE = 90 degrees

AB=BC=ED

AE=CD

Stephanie wants to maximise the area of the pentagon.

Use calculus to find the maximum area of the pentagon. [10 marks]

im not entirely how to approach the question. ive tried and ive got no where near the answer.
0
3 years ago
#2
(Original post by ZainulAli1234)
Stephanie is making a wire sculpture. She bends a piece of thin wire of length 48cm into a Pentagon ABCDE

Angle AED= Angle CDE = 90 degrees

AB=BC=ED

AE=CD

Stephanie wants to maximise the area of the pentagon.

Use calculus to find the maximum area of the pentagon. [10 marks]

im not entirely how to approach the question. ive tried and ive got no where near the answer.
What have you tried, exactly??

The main things to do here is to introduce a variable concerning the sides of the pentagon, and hence determine what the area is in terms of this variable. Having that, it's straight-forward maximisation problem from here. A diagram of this pentagon would also be useful to draw.
0
3 years ago
#3
Try drawing it out (marking out the sides of the same length, and the right angles), and seeing if you can spot how you would calculate the area. Since you know three sides are the same length, and another pair are, and their total length, you can get an expression for the total length in terms of one variable. Choosing your variable appropriately will let you write an expression for the area more easily too.
0
#4
there is a diagram its basically a rectangle but on one side instead of a side there are two lines making a point of a triangle.
so ive made AB=Bc=ED be 3x and AE=CD be 2y.I knew that since the perimeter was 48 => 3x+2y=48.
after this i thinking of using the 1/2absinc equation to find the area of the triangle bt its not calculus.
(Original post by RDKGames)
What have you tried, exactly??

The main things to do here is to introduce a variable concerning the sides of the pentagon, and hence determine what the area is in terms of this variable. Having that, it's straight-forward maximisation problem from here. A diagram of this pentagon would also be useful to draw.
0
3 years ago
#5
(Original post by ZainulAli1234)
there is a diagram its basically a rectangle but on one side instead of a side there are two lines making a point of a triangle.
so ive made AB=Bc=ED be 3x and AE=CD be 2y.I knew that since the perimeter was 48 => 3x+2y=48.
after this i thinking of using the 1/2absinc equation to find the area of the triangle bt its not calculus.
Yes, is an equation to be satisfied. Leave it alone for now.

First express the area in terms of . The rectangle is obviously, but for the triangle you can make use of the formula 1/2absinC. Hopefully the angle C is obvious, given that this triangle of interest is actually equilateral.

The calculus bit comes later, so for now just work out what the area of the pentagon is in terms of .
0
3 years ago
#6
Any chance you could post a photo of the question, please?
(Original post by ZainulAli1234)
there is a diagram its basically a rectangle but on one side instead of a side there are two lines making a point of a triangle.
so ive made AB=Bc=ED be 3x and AE=CD be 2y.I knew that since the perimeter was 48 => 3x+2y=48.
after this i thinking of using the 1/2absinc equation to find the area of the triangle bt its not calculus.
0
9 months ago
#7
(Original post by ZainulAli1234)
Stephanie is making a wire sculpture. She bends a piece of thin wire of length 48cm into a Pentagon ABCDE

Angle AED= Angle CDE = 90 degrees

AB=BC=ED

AE=CD

Stephanie wants to maximise the area of the pentagon.

Use calculus to find the maximum area of the pentagon. [10 marks]

im not entirely how to approach the question. ive tried and ive got no where near the answer.
3x + 2y = 48
y = 1/2(48-3x) = 24 - 1.5x

Area of rectangle = x(24-1.5x) = 24-1.5x^2
Area of equilateral triangle = 1/2absinC = 1/2x*xsin60 = (√3/4)x^2

Total Area = (√3/4)x^2 + 24 -1.5x^2
Simplified Area = ((√3-6)/4)x^2 +24x

dA/dx = ((√3-6)/2)x + 24
((√3-6)/2)x + 24 = 0
((√3-6)/2)x = -24
(√3-6)x = -48
x = (96 + 16√3x)/11

d^2A/dx^2 = (√3-6)/2
(√3-6)/2 < 0 therefore maximum

Area = 134.95cm^2
Area = 135cm^2 3sf.
1
9 months ago
#8
(Original post by Asuleman04)
3x + 2y = 48
You're replying to a two year old thread, but can you remove or edit your post please - you are breaking forum rules by posting a full solution!

Thanks 0
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