Asked by Anonymous
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
Answered by
Reiny
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.