Differential Calculus Volume

[JimCrown]

Independentlyapply differential calculus to the determination of maxima and minima inindustry-related problems.




Use differential calculus to determine minima values of arequired volume.



A rectangular sheet of metal having dimensions 30 cm by18 cm has the squares removed from each side of the four corners and the sidesbent upwards to form an open box.



Determine the maximum possible volume of the box, givingyour answer in the correct SI units and to 3 significant figures.




I have only seen these done for height on a graph. So, Itried and became stuck at a point. Can I have some help and be pointed in theright direction. I don’t know how to get the volume. Here is what I have done




Volume = Length x Width x Height

L = 30 W = 18
Area = L*W
= L*W – 4x
Area = 540 – 4x
Area = (18 – 2x) x(30 – 2x)
= (540 – 36x – 60x + 4x^2) * x
= (540 – 96x + 4x^2)
= (135 – 2x + x^2)
X * (x^2 – 2x + 135)
= x^3 – 2x^2 + 135x
Derivative:
= 3x^2 – 48x + 135
2^nd Derivative:
= 6x – 48x + 135
+48x
48x = 6x + 135
42x = 135
x = 3.214285714
4x = 12.85714286
540 – 4x = 527.1428571

Do I then times by 3.214285714 to get the maximum volume

 

[JeffM]

Independentlyapply differential calculus to the determination of maxima and minima inindustry-related problems.
I have only seen these done for height on a graph. So, I tried and became stuck at a point. Can I have some help and be pointed in theright direction. I don’t know how to get the volume. Here is what I have done

Volume = Length x Width x Height

That is correct. But it is not the length and width of the sheet. It is the length and width after the corner pieces are cut and the sides bent up.

L = 30 W = 18

Area = L*W
= L*W – 4x
Area = 540 – 4x

So that is wrong.

Area = (18 – 2x) x(30 – 2x)

= (540 – 36x – 60x + 4x^2) * x
= (540 – 96x + 4x^2)

This is correct.

= (135 – 2x + x^2)

This is wrong. Why are you dividing the area by 4? And how do you get 2 from dividing 96 by 4?

By the way. Avoid using X to indicate multiplication. It is easy to mix it up with x as a variable.

So let's do this correctly.

\(\displaystyle volume = v.\)

But volume = length * width * height AFTER the cuts are made.

\(\displaystyle length = 30 - 2x.\)

\(\displaystyle width = 18 - 2x.\)

\(\displaystyle height = x.\)

\(\displaystyle v = (30 - 2x)(18 - 2x)x = 540x - 96x^2 + 4x^3.\)

A volume is going to involve a cubic so you have a reasonableness check.

Now you take the derivative of v with respect to x.

\(\displaystyle \dfrac{dv}{dx} = 540 - 192x + 12x^2.\)

Now you set the derivative to zero and solve for x.

\(\displaystyle 540 - 192x + 12x^2 = 0 \implies x^2 - 16x + 45 = 0 \implies x^2 - 16x + 64 = 64 - 45 \implies\)

\(\displaystyle x - 8 = \pm \sqrt{19} \implies x = 8 \pm \sqrt{19} \approx 12.359 \text { or } 3.641.\)

But 12.359 is physically impossible so \(\displaystyle x \approx 3.641.\)

Let's check that gives a maximum.

\(\displaystyle x = 3 \implies v = (30 - 6)(18 - 6)3 = 864.\)

\(\displaystyle x = 4 \implies v = (30 - 8)(18 - 8)4 = 880.\)

\(\displaystyle x \approx 3.641 \implies v = (30 - 2 * 3.641)(18 - 2 * 3.641)3.641 \approx 886.553.\)