It depends on the shape of the 5400cm^2 cardboard.
If it is square, it is sqrt5400 or 73.4cm to a side.
Now set it down, call the sides s, and the height h.
You have to cut out h^2 from each of the four corners.
Area=b^2 h
but 2h+b=73.4 or h=(73.4-b)/2
Area=b^2(73.4-b)/2
dArea/db= 2b(73.4-b)/2 -b^2/2=0
solve for b
Then use that to go back and solve for h.
A box with a square base and an open top is constructed from 5400 cm^2 of cardboard. Find the dimensions of the largest possible box.
I know the answer is :
base lenght - 42.4 cm
height- 21.2 cm
please help me, thank you so much :)
2 answers
If this is the typical questions, identical squares are to be cut out of each of the corners of the cardboard, and then the sides folded up to form the box.
Let each side of the cutout be x cm
so each side of the base of the box will be √5400 - 2x
volume of box
= x(√5400-2x)(√5400-2x)
= 5400x-4√5400x^2 + 4x^3
d(volume)/dx = 5400 - 8√5400x + 12x^2 = 0 for a max of volume
x^2 - 20√6 + 450 = 0
using the quadratic formula
x = (20√6 ± √(2400 -4(1)(450))/2
= (20√6 ± √600)/2
= 36.74 or 12.25
The first answer gives us a volume of zero, the minimum, and the second gives us the maximum value
height = 12.25
base = √5400-2x = 48.99
Let each side of the cutout be x cm
so each side of the base of the box will be √5400 - 2x
volume of box
= x(√5400-2x)(√5400-2x)
= 5400x-4√5400x^2 + 4x^3
d(volume)/dx = 5400 - 8√5400x + 12x^2 = 0 for a max of volume
x^2 - 20√6 + 450 = 0
using the quadratic formula
x = (20√6 ± √(2400 -4(1)(450))/2
= (20√6 ± √600)/2
= 36.74 or 12.25
The first answer gives us a volume of zero, the minimum, and the second gives us the maximum value
height = 12.25
base = √5400-2x = 48.99
3 answers