Answer:
Yes
8 ft
Step-by-step explanation:
The equation of the arch is
[tex]y=-0.0625x^2+1.25x+5.75[/tex]
Differentiating with respect to [tex]x[/tex] we get
[tex]\dfrac{dy}{dx}=-2\times 0.0625x+1.25=-0.125x+1.25[/tex]
Equating with zero
[tex]0=-0.125x+1.25\\\Rightarrow x=\dfrac{-1.25}{-0.125}\\\Rightarrow x=10[/tex]
Double derivative of the equation
[tex]\dfrac{d^2y}{dx^2}=-0.125<0[/tex]
So, the function is maximum at [tex]x=10[/tex]
[tex]y=-0.0625x^2+1.25x+5.75=-0.0625\times 10^2+1.25\times 10+5.75\\\Rightarrow y=12[/tex]
The function is maximum at [tex](10,12)[/tex]
So, the truck's center should be below the point [tex](10,12)[/tex] for maximum width to pass through.
Now truck is 11 feet tall so [tex]y=11[/tex]
[tex]11=-0.0625x^2+1.25x+5.75\\\Rightarrow -0.0625x^2+1.25x-5.25=0\\\Rightarrow x=\frac{-1.25\pm \sqrt{1.25^2-4\left(-0.0625\right)\left(-5.25\right)}}{2\left(-0.0625\right)}\\\Rightarrow x=6,14[/tex]
The maximum width of the truck can be [tex]14-6=8\ \text{ft}[/tex] if it is 11 feet tall.
So, the truck which has a width of 7 feet can fit under the arch.
