Completely lost on this calculus optimization problem. Please help

Let α = angle of elevation from observers eyes to top of sign

Vertical distance from observer's eyes to top of sign = (50 - 5) + h = 45+h ft
Horizontal distance from observer's eyes to building = x

tan α = (45+h)/x

Let β = angle of elevation from observers eyes to bottom of sign

Vertical distance from observer's eyes to bottom of sign = 50 - 5 = 45 ft
Horizontal distance from observer's eyes to building = x

tan β = 45/x

Let θ = angle between the lines of sight from the observer's eye to the top and the bottom of the sign. Since the best view occurs when this angle is largest possible, θ is the variable we wish to maximize

θ = α - β
tan θ = (tan α - tan β) / (1 + tan α tan β)
tan θ = ((45+h)/x - 45/x) / (1 + (45+h)/x * 45/x)
tan θ = ((45 + h - 45)/x) / ((x² + 45(45+h))/x²)
tan θ = (h/x) / ((x² + 45h + 2025)/x²)
tan θ = hx / (x² + 45h + 2025)

θ = arctan [ hx / (x² + 45h + 2025) ]

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We want to show that when sign is 10 ft tall, ideal distance from building is 49.7 feet

First we differentiate θ with respect to x, keeping h constant.

θ = arctan [ hx / (x² + 45h + 2025) ]

dθ/dx = 1/{[hx/(x²+45h+2025)]² + 1} * (h(x²+45h+2025) - hx(2x))/(x²+45h+2025)²
dθ/dx = (x²+45h+2025)²/(h²x²+(x²+45h+2025)²) * h(x²+45h+2025-2x²)/(x²+45h+2025)²
dθ/dx = 1/(h²x² + (x²+45h+2025)²) * h (45h + 2025 - x²)
dθ/dx = h (45h + 2025 - x²) / (h²x² + (x²+45h+2025)²)

dθ/dx = 0
45h + 2025 - x² = 0
x² = 45h + 2025

We take positive root, since x < 0 would put us on the other side of the building

x = √(45h + 2025)

So for constant h, dθ/dx = 0 when x = √(45h + 2025)

Note that dθ/dx > 0 when 0 < x < √(45h + 2025)
and that dθ/dx < 0 when x > √(45h + 2025)
So θ increases from 0 to √(45h + 2025), and then decreases

Therefore, θ is maximized when x = √(45h + 2025) feet