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
2nd 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