Let us say that x is the cut that we will make on the
sides to make a box, therefore the new dimensions are:
l = 15 – 2x
w = 8 – 2x
It is 2x since we cut on two sides.
We know that volume is:
V = l w x
V = (15 – 2x) (8 – 2x) x
V = 120x – 30x^2 – 16x^2 + 4x^3
V = 120x – 46x^2 + 4x^3
Taking the 1st derivative:
dV/dx = 120 – 92x + 12x^2
Set dV/dx = 0 to get maxima:
120 – 92x + 12x^2 = 0
Divide by 12:
x^2 – (92/12)x + 10 = 0
(x – (92/24))^2 = -10 + (92/24)^2
x - 92/24 = ±2.17
x = 1.66, 6
We cannot have x = 6 because that will make our w
negative, so:
x = 1.66 inches
So the largest volume is:
V = 120x – 46x^2 + 4x^3
V = 120(1.66) – 46(1.66)^2 + 4(1.66)^3
V = 90.74 cubic inches
Answer:
0.500 T
Explanation:
Since the change in time and the number of coils are both 1, I set the problem up to be 1.3=(1.5(x)-13(x)). I then plugged in numbers for x until I got the answer to be 1.3 V.
Answer:
22.36 rad
Explanation:
Applying,
ω = θ/t.............. Equation 1
Where ω = angular velocity, θ = angular displacement of the baseball, t = time
make θ the subject of the equation
θ = ωt............... Equation 2
From the question,
Given: ω = 350 rev/min = 350(0.10472) = 36.652 rad/s, t = 0.61 s
Substitute these values into equation 1
θ = 0.61(36.652)
θ = 22.36 rad
Hence the angular displacement of the baseball is 22.36 rad