Question
AN ISOSCELES TRIANGLE IS TO HAVE A PERIMETER OF 64CM. DETERMINE THE SIDE LENGTHS OF THE TRIANGLE IF THE AREA IS TO BE A MAXIMUM.
Answers
Make a diagram of an isosceles triangle.
Let the base be 2x , the height is h, and each of the equal sides is y
then 2x+2y= 64
x+y = 32
y = 32-x (equation#1)
also x^2 + h^2 = y^2
h^2 = y^2 - x^2
= (32-x)^2 - x^2
= 1024 - 64x
h = √(1024-64x) = (1024-64x)^(1/2) (#2)
Area = (1/2)(2x)h
= xh
= x(1024-x)^(1/2)
Notice by letting the base be 2x , fractions were avoided so far.
Time for you to take over.
take the derivative of area using the product rule, set it equal to zero and solve for x
Sub into #2 to get h
Let the base be 2x , the height is h, and each of the equal sides is y
then 2x+2y= 64
x+y = 32
y = 32-x (equation#1)
also x^2 + h^2 = y^2
h^2 = y^2 - x^2
= (32-x)^2 - x^2
= 1024 - 64x
h = √(1024-64x) = (1024-64x)^(1/2) (#2)
Area = (1/2)(2x)h
= xh
= x(1024-x)^(1/2)
Notice by letting the base be 2x , fractions were avoided so far.
Time for you to take over.
take the derivative of area using the product rule, set it equal to zero and solve for x
Sub into #2 to get h
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