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.