the perimeter of an isosceles triangle is 10 cm. Find the dimensions of the tringle that produce a min. area

Question

i neep help asap

Answer 1

Ivan, you need to come up with an equation of the area of the triangle as a function of a single variable in which is related a length of a side. Start with the perimeter of the triangle in which let y = the triangle's base and x = the length of one of the sides of the triangle. Hence, the perimeter is

 

P=y + 2x = the perimeter of the isosceles triangle. This implies y=P-2x

 

Let h = the altitude of the triangle in which is:

 

h(x)=√(x^2-(1/2*y)^2)=√(x^2-1/4*y^2)=√(x^2-1/4*(P-2x)^2)

 

Thus, A(x) = 1/2 *y*h(x)=(P-2x)*√(x^2-1/4*(P-2x)^2) = the area of the triangle as a function of x.

 

Take the derivative of A(x) with respect to x and set it to equal to zero.

 

A'(x)=0=1/2*(dy/dx*h(x) + y*dh/dx) Then solve for x. This would give the dimensions of the triangle with maximum area. If x = 5cm, this would give the area being zero in which would be the minimum area. Also, with some algebraic manipulation for h(x), it becomes:

 

h(x)=√(Px-1/4P^2) in which makes x=1/4*P = 2.5cm to make h(x)=0

 

Perhaps, the problem is asking what are the dimensions of the isosceles triangle to give its maximum area.