find a rectangular contains at least 231 cubic inches.the height of the box must be exactly 1/4th of the width of the box. the box is closed on all 6 sides. the surface area of the box minimized.?
To find the dimensions of a rectangular box that contains at least 231 cubic inches and has a minimum surface area, we can solve this problem using calculus.
Let's assume the width of the box is "w" inches. Since the height of the box is 1/4th of the width, the height of the box will be (1/4)w inches. The length of the box is not given, so let's call it "l" inches.
The volume of a rectangular box is given by the formula: V = l * w * h
In this case, the volume of the box is at least 231 cubic inches, so we have the inequality: lw(1/4)w >= 231
Simplifying this inequality, we get: w^2 * l >= 924
Now, let's find the surface area of the box. The surface area is the sum of the areas of all six sides of the rectangular box.
The surface area of a rectangular box is given by the formula: S = 2lw + 2lh + 2wh
Substituting the values of height (h = (1/4)w) and simplifying, we get: S = 2lw + (3/2)w^2
Now, let's solve this problem using calculus.
1. Rewrite the surface area equation in terms of a single variable (w):
S = 2lw + (3/2)w^2
S = 2l(924 / l) + (3/2)w^2
S = 1848 / l + (3/2)w^2
2. Express the volume equation in terms of a single variable (w):
lw(1/4)w >= 231
(1/4)w^2l >= 231
l >= 924 / w^2
3. Substitute the value of "l" from step 2 into the surface area equation:
S = 1848 / (924 / w^2) + (3/2)w^2
S = 2w^2 + (3/2)w^2
S = (7/2)w^2
4. Take the derivative of the surface area equation with respect to "w" to find the critical points:
dS/dw = 0
(7/2)(2w) = 0
7w = 0
w = 0
5. Analyze the critical point found in step 4:
Since width cannot be zero, this critical point is not valid.
6. Determine the boundaries for "w":
Since the width must be positive, we can assume w > 0.
Also, from the volume equation, l >= 924 / w^2.
7. Determine the limit as w approaches infinity:
As w gets larger and larger, the surface area (S) also gets larger and larger, but it never reaches infinity.
8. Determine the limit as w approaches 0:
As w approaches 0, the surface area (S) also approaches 0.
9. Compare the surface area at the boundaries and critical points:
- As w approaches infinity, S approaches infinity
- As w approaches 0, S approaches 0
- Since there are no valid critical points, we can ignore that case.
10. Conclusion:
The minimum surface area occurs at one of the boundaries.
We need to find the value of "w" that minimizes the surface area.
To solve for the minimum surface area, we need to find the value of "w" that gives us the minimum surface area within the given boundary conditions.