Find the maximal area of right triangle with hypotenuse of length 1.
The area of a triangle is:
\[ A= \frac{bh}{2} \]
And given the hypotenuse we can use the Pythagrean Theorem:
\[ c^2 = a^2 + b^2 \]
To find the maximum value of this shape we use the Pythagream Theorm in the Area function and get the following functions. We know that c is equal to 1 therefore:
\[ a = \sqrt{1-b^2} \]
And the area function can now be written as:
\[ A = \frac{\sqrt{1-b^2}*b}{2} \] To sovle this problem we need to find the derivative of the above formula. Using R we get the following result.
f = expression((sqrt(1-b^2)*b)/2)
cat("\nUsing D() function:\n")##
## Using D() function:print(D(f, 'b'))## (sqrt(1 - b^2) - 0.5 * (2 * b * (1 - b^2)^-0.5) * b)/2\[ -\frac{2b^2 - 1}{2\sqrt{1-b^2}} \]
The zeros can be found at :
\[ b = \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \] Where the maximum would be the positive value.
b <- 1/sqrt(2)
eval(f,b)## [1] 0.25By substituting the value found when the derivative is equal to 0, the maximum area is 1/4.