ASAP help needed on MAX/MIN questions (1 Viewer)

[Petinga]

1. A rectangular box whose base is square is to be made so that its total surface area is constant. Prove that the volume of the box is greatest if the box is a cube.

2. Find the dimensions of the largest rectangle that can be inscribed in the semicircle y= route 4 - x squared

3. The runing cost (cost of fuel) for a certain ship is $3 per hour when the ship is not moving, and this cost increases by an amount that is proportional to the cube of its speed, V Km per hour. If the running cost per hour is $6.75 when the speed is 15 Km per hour, obtain a fromula for the runnig cost per hour at speed V, and calculate the value of V for which the total running cost for a journey of 45oKm is a minimum.


If anyone can help ASAP

 

[PtolemysPenguin]

1. A rectangular box whose base is square is to be made so that its total surface area is constant. Prove that the volume of the box is greatest if the box is a cube.

2. Find the dimensions of the largest rectangle that can be inscribed in the semicircle y= route 4 - x squared

3. The runing cost (cost of fuel) for a certain ship is $3 per hour when the ship is not moving, and this cost increases by an amount that is proportional to the cube of its speed, V Km per hour. If the running cost per hour is $6.75 when the speed is 15 Km per hour, obtain a fromula for the runnig cost per hour at speed V, and calculate the value of V for which the total running cost for a journey of 45oKm is a minimum.


If anyone can help ASAP

1. Let dimensions be x, x and y.
Surface area = 2x^2 + 4xy = c (a constant)
Therefore 4xy = c - 2x^2
y = c/4x - x/2
Volume (V) is (x^2)y = cx/4 - (x^3)/2
V' = c/4 - (3/2)x^2
When V' = 0, c/4 = (3/2)x^2
Therefore x^2 = c/6
Therefore x = (c/6)^(1/2) (x represents a length, and lengths are positive)
V'' = -(3/2)x, which is negative when x = (c/6)^(1/2), therefore x = (c/6)^(1/2) is an absolute maximum for V.

Now, 2x^2 + 4xy = c
x = (c/6)^(1/2), therefore 2x^2 = c/3
Therefore c/3 + 4xy = c
Therefore 4xy = 2c/3
xy = c/6
But x = (c/6)^(1/2)
Therefore y = (c/6)^(1/2) = x
Therefore y = x and the box is a cube.

I'll do 2 and 3 later