A box (with no top) is to be constructed from a piece of cardboard of sides A and B by cutting out squares of length h from the corners and folding up the sides. Find the value of h that maximizes the volume of the box if
A = 7 and B = 12
Draw a rectangle 7 x 12.
In the corners draw four squares of dimensions h x h.
It doesn't matter what the length of h is.
Dimensions of a new rectangle will be ( 7 - 2 h ) x ( 12 - 2 h )
The volume of the box will be:
Area of a new rectangle ∙ length h
V = ( 7 - 2 h ) ∙ ( 12 - h ) ∙ h
V = ( 84 - 24 h - 7 h + 2 h² ) ∙ h
V = ( 2 h² - 31 h + 84 ) ∙ h
V = 2 h³ - 31 h² + 84 h
The function has an exstrem (maximum or minimum) where the first derivative is zero.
In this case you need to find:
dV / dh = V′( h ) = 0
( 2 h³ - 31 h² + 84 h )′ = 2 ∙ 3 h² - 31 ∙ 2 h + 84 = 0
6 h² - 62 h + 84 = 0
The solutios are:
h = ( 31 -√457 ) / 6 ≈ 1.60374
and
h = ( 31 +√457 ) / 6 ≈ 8.7296
Now you have to second derivative test.
If V"( h ) < 0 then function has maximum at h.
If V" ( h ) > 0 then function has minimum at h.
V" ( h ) = 2 ∙ 6 h - 62
V" ( h ) = 12 h - 62
For
h = 1.60374
V" ( h ) = 12 ∙ 1.60374 - 62 = − 42.75512 < 0
For
h = 8.7296
V" ( h ) = 12 ∙ 8.7296 - 62 = 42.7552 > 0
So
For h = ( 31 -√457 ) / 6 ≈ 1.60374 volume has a maximum.
For h = ( 31 +√457 ) / 6 ≈ 8.7296 volume has a minimum.
By the way dimension h = 8.7296 are impossible because for h = 8.7296 dimension 7 - 2 h would have a negative value and the length cannot be negative.
Well, constructing a box sounds like a pretty square deal to me, but let's get down to business. To find the value of h that maximizes the volume of the box, we need to find the dimensions of the box in terms of h.
Let's call the length of the box x and the width y. When we cut squares of length h from the corners and fold up the sides, the length of the box will be reduced by 2h and the width will be reduced by 2h. So, we can express the dimensions of the box as:
x = A - 2h
y = B - 2h
Now, the volume of the box is given by V = x * y * h. Substituting the values of x and y in terms of h, we get:
V = (A - 2h) * (B - 2h) * h
Now, let's differentiate this volume equation with respect to h and find its critical points. Differentiating and simplifying, we get:
V' = -6h^2 + 38h - 84
Setting V' equal to zero to find the critical points:
-6h^2 + 38h - 84 = 0
Using the quadratic formula, we find two possible values for h: h ≈ 2.64 and h ≈ 5.69.
However, we need to check whether these values actually correspond to a maximum or minimum. Differentiating V' with respect to h again, we find:
V'' = -12h + 38
Substituting the values of our critical points, we get:
V''(2.64) ≈ 8.32
V''(5.69) ≈ -1.28
Since V''(2.64) > 0, this means h ≈ 2.64 corresponds to a local minimum and V''(5.69) < 0, which means h ≈ 5.69 corresponds to a local maximum.
So, the value of h that maximizes the volume of the box is approximately h ≈ 5.69.
To find the value of h that maximizes the volume of the box, we can start by determining the expression for the volume of the box in terms of h.
The length of the box will be A - 2h, and the width will be B - 2h (since squares of side length h are cut out from the corners).
The height of the box will be equal to h.
Therefore, the volume of the box is given by:
V = (A - 2h)(B - 2h)(h)
Plugging in the values A = 7 and B = 12, we have:
V = (7 - 2h)(12 - 2h)(h)
Now we can find the value of h that maximizes V.
First, let's expand the expression for V:
V = (84 - 19h + 4h^2)(h)
V = 4h^3 - 19h^2 + 84h
To find the maximum value of V, we can take the derivative with respect to h and set it equal to zero:
dV/dh = 12h^2 - 38h + 84
Setting this equal to zero and solving for h:
12h^2 - 38h + 84 = 0
Unfortunately, this equation does not factor easily. Therefore, we can use the quadratic formula to find the values of h:
h = (-b ± √(b^2 - 4ac))/2a
For our equation, a = 12, b = -38, and c = 84. Plugging in these values:
h = (-(-38) ± √((-38)^2 - 4(12)(84)))/(2(12))
Simplifying, we have:
h = (38 ± √(1444 - 4032))/24
h = (38 ± √(-2588))/24
Since we cannot take the square root of a negative number, there are no real solutions for h that maximize the volume of the box.
To find the value of h that maximizes the volume of the box, we need to express the volume of the box as a function of h, and then maximize that function.
Let's start by visualizing the situation. We have a rectangular piece of cardboard with sides A = 7 and B = 12. We will cut squares of side length h from each corner and fold up the sides to form an open-top box.
To maximize the volume of the box, we need to maximize the product of its dimensions. The length of the box will be (B - 2h), and the width will be (A - 2h) after folding up the sides.
The height of the box will be h. So, the volume of the box can be expressed as:
V(h) = h * (B - 2h) * (A - 2h)
To find the value of h that maximizes the volume of the box, we need to find the critical points of V(h). We can do this by finding the derivative of V(h) with respect to h and setting it equal to 0.
Let's find the derivative of V(h):
V'(h) = (B - 2h)(A - 2h) + h(-2)(A - 2h) + h(-2)(B - 2h)
V'(h) = (A - 2h)(B - 2h) - 2h(A - 2h) - 2h(B - 2h)
V'(h) = (A - 2h)(B - 2h) - 2h(A - 2h - B + 2h)
V'(h) = (A - 2h)(B - 2h) - 2h(A - B)
Now, set V'(h) = 0 to find the critical points:
(A - 2h)(B - 2h) - 2h(A - B) = 0
Expand and simplify the equation:
AB - 2hA - 2hB + 4h^2 - 2hA + 2hB = 0
4h^2 - 4hA + AB = 0
Divide the entire equation by 4:
h^2 - hA + (AB/4) = 0
Now, we have a quadratic equation in terms of h. We can solve this equation to find the value(s) of h that maximize the volume of the box.
Using the quadratic formula, we have:
h = (-(-A) ± √((-A)^2 - 4(1)(AB/4))) / (2(1))
Simplifying further:
h = (A ± √(A^2 - AB)) / 2
Substituting A = 7 and B = 12 into the equation, we have:
h = (7 ± √(7^2 - (7)(12))) / 2
h = (7 ± √(49 - 84)) / 2
h = (7 ± √(-35)) / 2
Since we cannot have a negative square root, the term inside the square root is negative, which means there are no real solutions for h in this case. Therefore, there is no value of h that maximizes the volume of the box.
This implies that the volume of the box will increase as h increases up to a certain point and then start decreasing again. However, since there are no real solutions, there is no specific value of h that maximizes the volume in this case.