You are given a 60 inch by 30 inch piece of cardboad and asked to make a six-sided box. If the cardboard is cut along the lines shown, what will the dimensions of the box with maximum volume?
3 answers
It would help if you would describe how the lines are drawn.
it is a rectangle with a shaded square in the top left and bottom left corners.....there is a shaded rectangle in the top right corner and bottom right corner....there is a shaded square to the left of the rectangle in the top right corner and the bottom right corner....in between the two shaded squares is an unshaded rectangle, both top and bottom....there are 2 horizonal lines that split the rectangle in 3 parts and there are 4 vertical lines that split the rectangle into four parts.
If you name the sides of the squares as of length x, then the horizontal rectangles are 30-x long, and the vertical ones are 30-2x long.
The total volume is therefore
A(x)=x(30-x)(30-2x)
which is a function of a single variable x.
Find the derivative of A(x) as A'(X) and equate A'(x)=0 to find the maximum area.
There will be two roots, x1 and x2. To distinguish the maximum from the minimum, we calculate A"(x) (second derivative of A(x)).
If A"(x)>0, it is a minimum.
If A"(x)<0, it is a maximum.
You will find that the smaller root is the maximum and equal to about 6.34, and the corresponding maximum volume is about 2598 in³.
The total volume is therefore
A(x)=x(30-x)(30-2x)
which is a function of a single variable x.
Find the derivative of A(x) as A'(X) and equate A'(x)=0 to find the maximum area.
There will be two roots, x1 and x2. To distinguish the maximum from the minimum, we calculate A"(x) (second derivative of A(x)).
If A"(x)>0, it is a minimum.
If A"(x)<0, it is a maximum.
You will find that the smaller root is the maximum and equal to about 6.34, and the corresponding maximum volume is about 2598 in³.
2 answers