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For the calculus lovers! An isosceles triangle has perimeter of 64 cm. determine the...

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For the calculus lovers! An isosceles triangle has perimeter of 64 cm. determine the side lengths if the triangle area is to be a maximum! HELP

Its university level calculus. I understand what's going on....u have three variables, and three formulas (perimeter, area of triangle, and pythagorean) you need to subsitute for 1 common variable and subsitute it into the surface are equation then find the derivative and it equate it to 0.

i get lost in the process

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Posted (Answer #1)

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The perimeter of the isosceles triangle is 2a+b =64 given.

Therefore, one of the isosceles side,a=(64-b)/2.

The height of the isosceles triangle from the base b is = sqrt(a^2-(b/2)^2) =sqrt{[(64-b)/2]^2-(b/2)^2}

=(1/2)sqrt{(64-b +b)(64-b-b)} =4sqrt(64-2b)

Therefore, the area of the triangle,A =(1/2)(Base)(height)=(b/2)(4)sqrt(64-2b)=2*b sqrt(64-2b)

By calculus  A is maximum if dA/db = 0 and d2A/db2 =-ve.

dA/db=0 =2*b*(-2)/[2sqrt(64-2b)]+2sqrt(64-2b)

=[-2b+2(64-2b]/sqrt(64-2b)= 2(64-3b)/sqrt(64-2b)=>

-3b=-64

b = 64/3

(d2A/db2] at b=64/3   is  2(64-2b)(-2b)/sqrt(64-2b)^(3/2) -4/sqrt(64-2b)  = -ve+-ve =-ve, as 64-2b is +ve

Threfore, the area of the isosceles triangle is highest for

b=64/3 and a =(64-64/3)/2 = 64/3 .

In other words, the area is maximum when the isosceles triangle becomes equilateral with each side measuring 64/3 cm.

sciencesolve's profile pic

Posted (Answer #2)

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The perimeter of any geometric shape may be found adding the lengths of the sides of the geometric shape.

Since the isosceles triangle has two equal sides the perimeter is evaluated such that:

P = 2x + y

The perimeter is given such that 64 = 2x + y.

The area of the isosceles triangle is: `A = y*height/2`

The height may be found using the Pythagora's theorem:

`x^2 = y^2/2 ` + height

height = `x^2 - y^2/2`

Use the relation involving perimeter to write y in terms of x.

y = 64 - 2x

height =`x^2 - (64 - 2x)^2/2`

height = `x^2 - 4(32 - x)^2/2 =gt`  height =`x^2 - 4(32 - x)^2`

Area `= ((64 - 2x)*(x^2 - 4(32 - x)^2))/2`

Area `= (2*(32 - x)*(x^2 - 4(32 - x)^2))/2`

Area `= (32 - x)*(x^2 - 4(32 - x)^2))`

Expanding the binomial yields:

`A(x) =(32 - x)*(x^2 - 4*32^2+64*4*x - 4x^2)`

`A(x) =(32 - x)*(-3x^2 - 4*32^2+64*4*x)`

`A(x) =(32 - x)*(x^2 -4096+256x - 4x^2)`

`A(x) =32x^2 -131072+8192x - 128x^2 - x^3 + 4096x - 256x^2 + 4x^3`

`` `A(x) = 3x^3 - 352x^2 + 12288x - 131072`

The area is given by the equation `A(x) = 3x^3 - 352x^2 + 12288x - 131072.`

The Area is maximum if A'(x) = 0.

You need to differentiate the function of area:

`A'(x) = 9x^2 - 704x + 12288`

If `A'(x) = 0 =gt 9x^2 - 704x + 12288 = 0`

Applying quadratic formula yields: `x_(1,2) = (704+-sqrt(495616 - 442368))/18`

`` `x_(1,2) ~~ (704+-230)/18`

`x_1~~ 51.88 ; x_2~~ 26.33`

The area is maximum if the lengths of the sides of the isosceles triangle are: `x_1~~ 51.88 ; x_2~~ 26.33.`

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