Differentiation - stationary points (20cm wire is bent to form isosceles triangle)

[sojeee]

A 20cm piece of wire is bent to form an isosceles triangle with base b. (So I labelled the two congruent sides as x each)

a) show that the area of the triangle is given by a = 1/2 x sqrt( 100b^2 - 10b^3)

my working:
p = 2x+b
20= 2x+b
x = (20- b)/2

x^2 = h^2 + (b/2)^2
x^2 - (b^2/4) = h^2

(20-b/2)^2 - (b^2/4) = h^2
(b^2 - 40b +400)/4 - b^2/4 = h^2
-40b + 400)/4 = h^2
h = sqrt (100-10b)

have I made a mistake somewhere ?

 

[Dr.Peterson]

A 20cm piece of wire is bent to form an isosceles triangle with base b. (So I labelled the two congruent sides as x each)

a) show that the area of the triangle is given by a = 1/2 x sqrt( 100b^2 - 10b^3)

my working:
p = 2x+b
20= 2x+b
x = (20- b)/2

x^2 = h^2 + (b/2)^2
x^2 - (b^2/4) = h^2

((20-b)/2)^2 - (b^2/4) = h^2
(b^2 - 40b +400)/4 - b^2/4 = h^2
-40b + 400)/4 = h^2
h = sqrt (100-10b)

have I made a mistake somewhere ?

At first I thought you had, until I saw you omitted the parentheses you had included earlier.

But your expression for h is good. Now use it to find the area.

And avoid using "x" for multiplication!

 

[sojeee]

At first I thought you had, until I saw you omitted the parentheses you had included earlier.

But your expression for h is good. Now use it to find the area.

And avoid using "x" for multiplication!

A = 1/2 * b * sqrt(100-10b)
I don’t know how to change the answer to look match theirs

 

[Dr.Peterson]

A = 1/2 * b * sqrt(100-10b)
I don’t know how to change the answer to look match theirs

The process you need is the opposite of simplifying a radical. There, you remove a perfect square from inside the radical, taking its square root as you do so. Here, you move a factor inside the radical, squaring it as you do so:

A = 1/2 b sqrt(100-10b) = 1/2 sqrt(b^2) sqrt(100-10b) = 1/2 sqrt(b^2 (100-10b)) = ...