Question

A truck that is 11 feet tall and 7 feet wide is traveling under an arch. The arch can be modeled by:

Answer 1

Answer:

Yes

8 ft

Step-by-step explanation:

The equation of the arch is

$y=-0.0625x^2+1.25x+5.75$

Differentiating with respect to $x$ we get

$\dfrac{dy}{dx}=-2\times 0.0625x+1.25=-0.125x+1.25$

Equating with zero

$0=-0.125x+1.25\\\Rightarrow x=\dfrac{-1.25}{-0.125}\\\Rightarrow x=10$

Double derivative of the equation

$\dfrac{d^2y}{dx^2}=-0.125$

So, the function is maximum at $x=10$

$y=-0.0625x^2+1.25x+5.75=-0.0625\times 10^2+1.25\times 10+5.75\\\Rightarrow y=12$

The function is maximum at $(10,12)$

So, the truck's center should be below the point $(10,12)$ for maximum width to pass through.

Now truck is 11 feet tall so $y=11$

$11=-0.0625x^2+1.25x+5.75\\\Rightarrow -0.0625x^2+1.25x-5.25=0\\\Rightarrow x=\frac{-1.25\pm \sqrt{1.25^2-4\left(-0.0625\right)\left(-5.25\right)}}{2\left(-0.0625\right)}\\\Rightarrow x=6,14$

The maximum width of the truck can be $14-6=8\ \text{ft}$ if it is 11 feet tall.

So, the truck which has a width of 7 feet can fit under the arch.