Calculus question need help pls?

Question

A rectangular block of metal with a square cross-section has a total surface area of 240cm^2. Find the maximum volume of the block?

To answer the question you need to use differentiation,
1) work out the equation for the volume?
which i think is x^2y (as area of cross section * length)
2) work out the equation for the surface area?
which i think is 2x^2+4xy = 240cm^2

i don't know what to do next to work out the values of x and y and find the 1st and 2nd derivatives and turning points etc...

could u pls help and show workings step by step pls.

Answer 1

First lets say the block has dimensions x, x, y
Now we need a formula for volume and a formula for S.A.
V = x^2y
S.A. = 2x^2 + 4xy = 240
We can rearrange the second to get y in terms of x
So 4xy = 240 - 2x^2 which implies y = (240 - 2x^2)/4x = (120 - x^2)/2x

We no substitute this into the first equation So V = x^2(120-x^2)/2x

So dividing top and bottom by x gives V = x(120 - x^2)/2 = 60x - x^3/2

To find the max we must differentiate and then let it equal zero
dV/dx = 60 - 3x^2/2

Let dV/dx = 0 implies 3x^2/2 = 60 So x^2 = 40 so x = root(40)
= 2root 10

To see if it is a max we must differentiate again

d2v/dx2 = -3x which is negitive for all positive x there x = root(40) is a max.

No sub this into our formula for V

V = 60x - x^3/2

So V = 60root(40) - 20root(40) = 40root(40) = 80root(10)

Answer 2

1. Let x be the length of the rectangle, and y be its width (and also the side of the square cross section.

The surface area is 4xy + 2y^2 = 240.
Let's resolve for x:
x=(240 - 2y^2)/4y = (60/y - y/2)

The volume is the area of the base * height, which is y^2 * x.
Let's call it v.

v = (60/y - y/2) * y^2 = 60y - y^3/2

dv/dy = 60 - 3y^2/2

At minimum and maximum points, the derivative is zero, hence we have the quadratic equation:

120 - 3y^2 = 0

y^2 = 40
or y=sqrt(40) [We do not take a negative value for measure].

x = 60/y - y/2 = 60/sqrt(40) + sqrt(40)/2 = 60*sqrt40/ 40 + sqrt(40)/2 = 3*sqrt(40/2 + sqrt(40)/2 = sqrt(40)

Now let's check if it is a minimum or a maximum;

d^2v/dy^2 v = d/dy (60 - 3y^2/2) = -6y = -6sqrt(4) < 0.
The cube's volume is the maximal.

The maximal volume is sqrt(40)^3 = sqrt(40)*sqrt(40)^2 = 40sqrt(40).

Answer 3

Well solve the equation

2*x^2 + 4*x*y = 240

for y:

y = (240 - 2*x^2)/(4*x)

and substitute in the formula for volume

V = x^2 * (240 - 2*x^2)/(4*x) = x * (240 - 2*x^2)/4

V = (240*x - 2*x^3)/4

Now you have a function of 1 argument and you differentiate it:

V' = (240 - 6*x^2)/4

V'' = -12*x/4 = -3*x < 0

The second derivative is always negative, so any critical point is a maximum. We find it for V' = 0

240 - 6*x^2 = 0 => x^2 = 40 => x = 2*sqrt(10)

The maximal volume is

Vmax = 2*sqrt(10)*(240 - 2*40)/4 = 80*sqrt(10)

By the way, this is a type of constrained optimization problem for a function of two variables. I used the elementary method of elimination because the constrain could be solved in terms of y.

Answer 4

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