Line segment 19 units long running from (x, 0) ti (0, y) show the area of the triangle enclosed by the segment is largest when x=y

The area of the triangle is

A = (xy) / 2

Also,

sqrt (x^2 + y^2) = 19

We solve this for y.

x^2 + y^2 = 361

y^2 = 361 - x^2

y = sqrt (361 - x^2)

Now we substitute this expression for y in the area equation.

A = (1/2) (x) (sqrt (361 - x^2))

A = (1/2) (x) (361 - x^2) ^ (1/2)

We take the derivative of A with respect to x.

dA/dx = (1/2) [ (x) * d/dx (361 - x^2) ^ (1/2) + (361 - x^2) ^ (1/2) ]

dA/dx = (1/2) [ (x) * (1/2) (361 - x^2) ^ (-1/2) (-2x) + (361 - x^2) ^ (1/2) ]

dA/dx = (1/2) [ (361 - x^2) ^ (-1/2) (-x^2) + (361 - x^2) ^ (1/2) ]

dA/dx = (1/2) [ (-x^2) / (361 - x^2) ^ (1/2) + (361 - x^2) / (361 - x^2) ^ (1/2) ]

dA/dx = (1/2) [ (-x^2 - x^2 + 361) / (361 - x^2) ^ (1/2) ]

dA/dx = (-2x^2 + 361) / [2 (361 - x^2) ^ (1/2) ]

Now we set the derivative equal to zero.

(-2x^2 + 361) / [2 (361 - x^2) ^ (1/2) ] = 0

-2x^2 + 361 = 0

-2x^2 = - 361

2x^2 = 361

x^2 = 361/2

x = 19/sqrt (2)

x^2 + y^2 = 361

(19/sqrt (2)) ^2 + y^2 = 361

361/2 + y^2 = 361

y^2 = 361/2

y = 19/sqrt (2)

We have maximum area at x = 19/sqrt (2) and y = 19/sqrt (2), or when x = y.