When h = 10, θ is maximized when x = √(45*10 + 2025) = √2475 = 49.7 ft
Good, our result matches what we were expecting!
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From above, we found that θ is maximized when x = √(45h + 2025) feet
To determine the height of the new sign so that the best view occurs when the observer is 60 feet from the building, let x = 60 then find value of h that maximizes θ
√(45h + 2025) = 60
45h + 2025 = 3600
45h = 1575
h = 35
Sign should be 35 feet tall for best view at 60 feet from building.
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To determine the height of the new sign so that the best view occurs when the observer is 35 feet from the building, let x = 35 then find value of h that maximizes θ
√(45h + 2025) = 35
45h + 2025 = 1225
45h = −800
h = −17.78
You cannot erect sign of any height above building so that best view is 35 feet from building. What h < 0 means, however, is that you CAN erect a sign hanging 17.78 ft from top of building (instead of on top of building) to get best view at 35 feet from building.
Mαthmφm