A lidless box with square ends is to be made from a thin sheet of metal. Determine the least area of the metal for which the volume of the box is 3.5 m^3.
If the box has ends of side x and length y, then x^2 y = 7/2
the area is a = 2x^2 + 4xy = 2x^2 + 4x(7/2x^2) = 2x^2 + 14/x
da/dx = 4x - 14/x^2
Now just find where da/dx=0
Well, I have to say, making a lidless box from a thin sheet of metal sounds like quite the unorthodox approach. It's like playing hide and seek without the hide part. But hey, who am I to judge?
To find the least area of the metal required, let's break it down. We know that the box has square ends, so let's call the length of each side x. The height of the box will be 3.5 m^3 divided by the area of the square base, which is simply x squared.
Now, to calculate the area of the metal, we need to consider that the box has two square ends and four sides. The area of each end will be x squared, while the area of each side will be x times the height.
So, the total area of the metal required will be:
2(x squared) + 4(x times height)
Substituting the height with 3.5 m^3 divided by x squared, we get:
2(x squared) + 4(x times (3.5 m^3 / x squared))
Now, to minimize the area, we can take the derivative with respect to x, set it equal to zero, and solve for x. But beware, calculus is a slippery slope, so be sure to have your clown shoes on!
Once you find the value of x that minimizes the area, you can plug it back into the equation to find the least area of the metal required. Happy calculating, my friend!
To determine the least area of the metal needed to construct the box, we need to optimize the dimensions of the box. Let's assume the side length of the square base of the box is x, and the height of the box is h.
Given that the volume of the box is 3.5 m^3, we can set up the following equation:
x^2 * h = 3.5
Now, we need to find the minimum surface area of the box. The surface area of the box consists of the area of the square base and the area of the four sides. Since the box is lidless, we don't need to consider the lid.
The area of the square base is given by:
A_base = x^2
The area of each side is given by:
A_side = x * h
We have four sides, so the total area of the four sides is:
4 * A_side = 4 * (x * h) = 4xh
Therefore, the total surface area, A_total, of the box is:
A_total = A_base + 4xh = x^2 + 4xh
To minimize the surface area, we need to minimize the value of A_total.
Now, we will solve for h in terms of x using the equation x^2 * h = 3.5:
h = 3.5 / x^2
Substituting this value of h into the equation for A_total, we get:
A_total = x^2 + 4xh = x^2 + 4x(3.5 / x^2) = x^2 + 14 / x
To find the minimum value of A_total, we can take its derivative with respect to x and set it equal to zero:
dA_total / dx = 2x - 14 / x^2 = 0
Multiplying through by x^2 to eliminate the denominator, we get:
2x^3 - 14 = 0
Simplifying further:
2x^3 = 14
x^3 = 7
Taking the cube root of both sides, we find:
x = ∛7
Therefore, the optimal value for x is the cube root of 7.
Finally, to find the minimum surface area, we substitute this value of x back into the equation for A_total:
A_total = x^2 + 14 / x = (∛7)^2 + 14 / (∛7) = 7^(2/3) + 14 * 7^(-1/3)
Calculating the numerical value, we can find the least area of the metal needed to construct the box.
To determine the least area of the metal required for the box, we need to minimize the surface area of the box.
Let's denote the width of the box as w and the height as h. Since the box has square ends, the length of the box would be the same as the width, so the length can also be denoted as w.
The volume of the box is given as 3.5 m^3, so we have the equation:
Volume = Length * Width * Height
3.5 = w * w * h
We want to minimize the surface area, so we need to express the surface area in terms of one variable (either w or h). The surface area of the box is given by:
Surface Area = 2[(Length * Width) + (Length * Height) + (Width * Height)]
In our case, since the ends of the box are square, the first term simplifies to 2(w * w), and the second and third terms simplify to 2(w * h):
Surface Area = 2(w * w) + 2(w * h) + 2(w * h)
Surface Area = 2w^2 + 4wh
Now, we can express the height (h) in terms of w using the volume equation:
3.5 = w * w * h
h = 3.5 / (w^2)
Substituting this expression for h in terms of w into the surface area equation:
Surface Area = 2w^2 + 4w(3.5 / w^2)
Surface Area = 2w^2 + 14 / w
Now, we need to find the value of w that minimizes the surface area. We can do this by taking the derivative of the surface area equation with respect to w and setting it equal to zero:
d(Surface Area) / dw = 4w - 14 / w^2
0 = 4w - 14 / w^2
To solve for w, we can multiply through by w^2:
0 = 4w^3 - 14
Now, we can solve this equation for w. By factoring, we can rewrite it as:
0 = (2w - √7)(2w^2 + √7w + √7)
Setting each factor equal to zero, we find two possible values for w:
2w - √7 = 0 => w = √7 / 2
2w^2 + √7w + √7 = 0 => no real solutions
Since width (w) cannot be negative, we discard the negative root. Therefore, the value of w that minimizes the surface area is w = √7 / 2.
To find the corresponding value of h, we substitute this value for w into the expression for h:
h = 3.5 / (w^2)
h = 3.5 / (√7 / 2)^2
h = 14 / 7
h = 2
Therefore, the width (w) is √7 / 2 and the height (h) is 2.
Finally, to find the least area of the metal, we substitute these values into the surface area equation:
Surface Area = 2w^2 + 4wh
Surface Area = 2(√7 / 2)^2 + 4(√7 / 2)(2)
Surface Area = 2 * 7 / 4 + 4 * 2 * √7 / 2
Surface Area = 7/2 + 8√7
So, the least area of the metal required for the box with a volume of 3.5 m^3 is 7/2 + 8√7 square units.