SOLUTION: Hi I would like a calculus answer to this problem (Using a derivative) What is the maximum area of a right angled triangle with perimeter 72? thank you!

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Question 894103: Hi I would like a calculus answer to this problem (Using a derivative)
What is the maximum area of a right angled triangle with perimeter 72?
thank you!
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Form a right triangle, sides A and B, hypotenuse C.
Perimeter:
A%2BB%2BC=72
Pythagorean theorem:
A%5E2%2BB%5E2=C%5E2
Let's get rid of C
C=72-A-B
C%5E2=%2872-A-B%29%5E2
.
.
A%5E2%2BB%5E2=A%5E2%2B2AB-144A%2BB%5E2-144B%2B5184
0=2AB-144A-144B%2B5184
B%282A-144%29=144A-5184
B=%28144A-5184%29%2F%282A-144%29
B=%2872%28A-36%29%29%2F%28A-72%29
Area (Use Y instead of A):
Y=%281%2F2%29AB
Maximize the area.
Make Y a function of only 1 variable.
Substitute,
Y=%281%2F2%29A%2A%28%2872%28A-36%29%29%2F%28A-72%29%29
Y=%2836A%5E2-1296A%29%2F%28A-72%29
Take the derivative using the quotient rule,
dY%2Fdx=%28%28A-72%29%2872A-1296%29-%2836A%5E2-1296A%29%281%29%29%2F%28A-72%29%5E2
dY%2Fdx=%2836%28A%5E2-144A%2B2592%29%29%2F%28A-72%29%5E2
Now set it equal to zero.
A%5E2-144A%2B2592=0
A%5E2-144A%2B5184%2B2592=5184
%28A-72%29%5E2=5184-2592
%28A-72%29%5E2=2592
A-72=0+%2B-+36sqrt%282%29
A=72+%2B-+36sqrt%282%29
Since A cannot be greater than 72,
A=72-36sqrt%282%29
Then,
B=%2872%28A-36%29%29%2F%28A-72%29
B=%2872%2836-36sqrt%282%29%29%29%2F%28-36sqrt%282%29%29
B=72-36sqrt%282%29
It's an isosceles right triangle.
So then,
Y=%281%2F2%29%28AB%29
Y=%281%2F2%29%2872-36sqrt%282%29%29%5E2
Y=%281%2F2%29%285184-2592sqrt%282%29-2592sqrt%282%29%2B2592%29
Y=%281%2F2%29%287776-5184sqrt%282%29%29
highlight%28Y=3888-2592sqrt%282%29%29