Respuesta :
Answer:
V = 56 mph.
The mileage of vehicle is 32 mi/gal
Explanation:
Given that
M (s) = - 1/28 v^2 + 4y - 80
But in the above equation on the place of y it should be V(velocity).
[tex]M(s)=-\dfrac{1}{28}V^2+4V-80[/tex] Â Â Â ---------1
We know that if we want to maximize the function then we differentiate that function .
So
[tex]\dfrac{dM}{dV}=-2\times \dfrac{1}{28}V+4[/tex]
[tex]\dfrac{dM}{dV}=- \dfrac{1}{14}V+4[/tex]
Foe maximum condition
[tex]\dfrac{dM}{dV}=0[/tex]
[tex]-\dfrac{1}{14}V+4=0[/tex]
So
V= 4 x 14
V = 56 mph.
So the speed of vehicle at best mileage is 56 mph.
Now by putting the value of V in the first equation
[tex]M(s)=-\dfrac{1}{28}V^2+4V-80[/tex] Â Â
[tex]M(s)=-\dfrac{1}{28}\times 56^2+4\times 56-80[/tex] Â Â
M(s) = 32
So the mileage of vehicle is 32 mi/gal
Answer:
32 mi/gal
Explanation:
[tex]M= -\frac{1}{28}v^{2}+4v-80[/tex] .... (1)
For best mileage, first find the derivative of mileage function with respect to v.
[tex]\frac{dM}{dv}=-\frac{2}{28}v+4[/tex]
Put it equal to zero.
[tex]\frac{1}{14}v = 4[/tex}
v = 56 mph
Put v = 56 in equation (1)
[tex]M= -\frac{1}{28}\times 56^{2}+4\times 56-80[/tex]
M = 32 mi/gal
