An open box is to be formed from a square sheet of carboard (square is 10x10 cm) by cutting squares and then folding up the sides. (the squares cut off are just the corners as they are labeles as an x by x).
A) Find a function for the volume of the box. State the domain of your function.
B)Find the value of x that maximizes the volume.
Base of the box would be (10-2x) by (10-2x) and the height would be x
a)
V = x(10-2x)(10-2x) , where 0 < x < 5
b) V = x(100 - 40x + 4x^2)
= 100x - 40x^2 + 4x^3
dV/dx = 100 - 80x + 12x^2
= 0 for a max of V
12x^2 - 80x + 100=0
3x^2 - 20x + 25 = 0
(x-5)(3x - 5) = 0
x = 5/3 or x = 5, but x ≠ 5 according to our domain
x = 5/3 will maximize the volume
i don't understand what is happening on the third line of B
I assumed you are studying Calculus, since this is a typical Calculus question.
Tell me otherwise
Im only in Precalc
can you just tell me how you got to dV/dx = 100-80x-12^2
I took the derivative of the function, which is one of the basic concepts in the Calculus course.
If you are not yet at that point, then this question should not have been put to you, since there is no other practical way to do this question.
I suppose you could use a "trial -and-error" method, that is, take different values of x in the above domain and see what kind of volume you would get.
e.g.
let x = 1
V = 1(8)(8) = 64
let x = 2
V = 2(6)(6) = 72 , which is more than 64
let x = 3
V = 3(4)(4) = 48 , ahh smaller
so lets try x = 2.5
V = 2.5(5)(5) = 62.5
how about 1.75
V = 1.75(6.5)(6.5) = 73.93 , heh, that's the biggest so far
how about 5/3 or 1.66666..
v = (5/3)(20/3)(20/3) = 74.074... ahh, even bigger
how about 1.65 , which is slighly < 5/3
v = 1.65(6.7)(6.7) = 74.0685 , a bit less
and finally how about x = 1.67 , a bit more than 5/3
V = 1.67(6.66)(6.66) = 74.0738.. , ahhh, a bit less than above
looks like my answer of x = 5/3 is right
To find the function for the volume of the box, we need to first determine the dimensions of the box in terms of the side length, x, that is cut off from each corner.
A) Volume of the box:
The base of the box will have dimensions (10 - 2x) by (10 - 2x). The height of the box will be x. Therefore, the volume of the box can be calculated as the product of the base area and the height:
Volume = (10 - 2x)(10 - 2x)(x)
Simplifying this expression gives:
Volume = 4x^3 - 60x^2 + 200x
The domain of this function is limited by the constraints imposed by the problem. Since we are cutting squares with side length x from each corner, the maximum value of x cannot exceed half the side length of the original square, which is 10/2 = 5 cm. Additionally, x cannot be negative or zero, as there would be no material left to form a box. Therefore, the domain of the function is 0 < x ≤ 5.
B) Finding the value of x that maximizes the volume:
To find the value of x that maximizes the volume, we need to find the critical points of the function and determine whether they correspond to maximum or minimum values.
1. Take the derivative of the volume function with respect to x:
dV/dx = 12x^2 - 120x + 200
2. Set the derivative equal to zero and solve for x to find the critical points:
12x^2 - 120x + 200 = 0
This is a quadratic equation which can be solved using the quadratic formula:
x = (-(-120) ± sqrt((-120)^2 - 4(12)(200))) / (2(12))
x = (120 ± sqrt(14400 - 9600)) / 24
x = (120 ± sqrt(4800)) / 24
x = (120 ± 2sqrt(1200)) / 24
x = (5 ± sqrt(300)) / 2
3. Simplify the expression for the critical points:
x1 = (5 + sqrt(300)) / 2
x2 = (5 - sqrt(300)) / 2
The critical points are x1 and x2.
4. Evaluate the second derivative of the volume function:
d²V/dx² = 24x - 120
5. Substitute the values of x1 and x2 into the second derivative:
d²V/dx² @ x1 = 24(5 + sqrt(300)) / 2 - 120 = 12sqrt(300) - 72
d²V/dx² @ x2 = 24(5 - sqrt(300)) / 2 - 120 = -12sqrt(300) - 72
6. Determine the nature of the critical points:
The nature of the critical points is determined by the sign of the second derivative. Positive second derivative indicates a minimum value, and negative second derivative indicates a maximum value.
Since x cannot be zero and sqrt(300) is positive, we have:
d²V/dx² @ x1 > 0, and d²V/dx² @ x2 < 0
Therefore, x1 corresponds to a minimum value, and x2 corresponds to a maximum value.
The value of x that maximizes the volume is x2:
x = (5 - sqrt(300)) / 2 ≈ 0.41 cm
Thus, cutting squares with a side length of approximately 0.41 cm from each corner will yield a box with maximum volume.