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youngstudent16
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Consider the semicircle with radius 1, the diameter is AB. Let C be a point on the semicircle and D the projection of C onto AB. Maximize the area of the triangle BDC.
What I'm thinking
y=sqrt(r^2-x^2) From the formula of a circle x^2+y^2=r^2
A=1/2(x+1)y The area of a triangle
A'=(1-x-2x^2)/2sqrt(1-x^2) Substitution from the formula of a circle and take the derivative
A'=0 if x=1/2 Set the derivative = to 0 to maximize it
y=sqrt(3)/2 Solve for y
A=1/2((1/2)+1)sqrt(3)/2= 3sqrt(3)/8 Plug in all the pieces into the original equation of the area of a triangle.
A=Area A'=Derivative of area
x+1 = base of the triangle
y=The heightThese are still new to me let me know how far off I am. I think I have the equation wrong.
(sorry posted this in wrong area first)
Consider the semicircle with radius 1, the diameter is AB. Let C be a point on the semicircle and D the projection of C onto AB. Maximize the area of the triangle BDC.
What I'm thinking
y=sqrt(r^2-x^2) From the formula of a circle x^2+y^2=r^2
A=1/2(x+1)y The area of a triangle
A'=(1-x-2x^2)/2sqrt(1-x^2) Substitution from the formula of a circle and take the derivative
A'=0 if x=1/2 Set the derivative = to 0 to maximize it
y=sqrt(3)/2 Solve for y
A=1/2((1/2)+1)sqrt(3)/2= 3sqrt(3)/8 Plug in all the pieces into the original equation of the area of a triangle.
A=Area A'=Derivative of area
x+1 = base of the triangle
y=The heightThese are still new to me let me know how far off I am. I think I have the equation wrong.
(sorry posted this in wrong area first)
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