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Kryger [21]
3 years ago
11

A cylinder is inscribed in a right circular cone of height 4.5 and radius (at the base) equal to 5. What are the dimensions of s

uch a cylinder which has maximum volume
Physics
1 answer:
ahrayia [7]3 years ago
4 0

Answer:

The maximum volume of inscribed cylinder  when its radius is r  =  10 \3 units  and its height is  1.5 units.

Explanation:

The formula to find the volume of inscribed cylinder is: V  =  (πr^2).y

let 'y' is the height of the cylinder and 'r' be the radius of the cylinder.

Radius of the inscribed cylinder is r = 5-x.

and the value of y is y = (4.5)/5 * x, it is find by the formula of slope because we know the slope of the slanted side of the cone. We go up 4.5 units for every 5 units we move to the right.

Now,

V = π(5-x)^2 × (4.5/5) . x

V = (4.5/5 . π) × (25x + x^3 - 10x^2)

And now to find the maximum volume derivative of V with respect to 'x' will be equal to zero like: dV/dx = 0

dV/dx = (4.5/5 . π) × (25 + 3x^2 - 20x)

(4.5/5 . π) × (25 + 3x^2 - 20x) = 0

25 + 3x^2 - 20x = 0

Write the equation in sequence and then use the factorization to find x

3x^2 - 20x + 25 = 0

3x^2 - 15x - 5x + 25 = 0

3x(x-5) -5(x-5) = 0

(x-5) (3x-5) = 0

x = 5/3 and x = 5

So, we have two values of 'x'. But the value 5/3 gives us the point of maximum by graphing V with 'x'. Because when the value is between 0 to 5 then the left edge of the inscribed cylinder stays properly within the cone.

Hence, at x = 5/3,

y = (4.5)/5 * x =  (4.5)/5 * 5/3

y = 1.5 ≈ 2.

and r = 5 - x = 5 - 5/3 = 10/3.

The maximum volume of inscribed cylinder  when its radius is r  =  10 \3 units  and its height is  1.5 units.

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A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist
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Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > 38  \frac{ft}{s^2}  x  \frac{1mi}{5280ft} x  \frac{(3600s)^2}{(1h)^2} = 93272.27  \frac{mi}{h^2}

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > v_2^2=v_1^2 + 2as

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Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

0^2 = 50^2 + 2(-93272.27)s&#10;&#10;0 = 2500 - 186544.54s&#10;&#10;Isolate S next.&#10;&#10;185644.54s= 2500&#10;&#10;s =  2500/(185644.54)&#10;&#10;s=0.0134&#10;

So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > 0.0134 mi *  \frac{5280ft}{1mi}  = 70.8 ft
So this means that the car traveled in feet 70.8 ft before it came to a stop.

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