Respuesta :
Answer:
(a) 30 feet
(b) 30 feet.
Step-by-step explanation:
Given that the opening of a tunnel that travels through a mountainside can be modeled by
[tex]y=\frac{-2}{15} (x-15)(x+15)[/tex], where x and y are measured in feet.
[tex]\Rightarrow y=\frac{-2}{15}(x^2-15^2}[/tex]
[tex]\Rightarrow y=\frac{-2}{15}{x^2-225}\cdots(i)[/tex]
(a) At the ground level, [tex]y=0[/tex]
So, the width of the tunnel at ground level is distance between the extreme point of the tunnel on the grount.
For, [tex]y=0[/tex], the extreme points of the tunnel.
[tex]0=\frac{-2}{15} (x-15)(x+15)[/tex]
[tex]\Rightarrow (x-15)(x+15) =0[/tex]
[tex]\Rightarrow x= 15, -15[/tex]
So, the extreme points of the tunnel are, [tex]x_1=15[/tex] and [tex]x_2=-15[/tex].
Hence, the width of the tunnel at the ground level
[tex]= | x_1 - x_2 |[/tex]
[tex]=|15-(-15)|[/tex]
[tex]=30[/tex] feet.
(b) The maximum height of the tunnel can be determiment by determining the maxima of the given function.
First determining the value of x for which the slope of the graph is zero.
[tex]\frac{dy}{dx}=0[/tex]
From equation (i),
-2x=0
[tex]\Rightarrow x=0[/tex]
And [tex]\frac{d^2y}{dx^2}= -2[/tex]
which is always negative, so at x=0 the value of y is maximum.
Again, put x=0 in equation (i), we have
[tex]y=\frac{-2}{15}{0^2-225}[/tex]
[tex]\Rightarrow y=30[/tex] feet.
Hence, the tunnel is 30 feet tall.
