A rectangular tank with a square base, an open top, and a volume of 864 ft^3 is to be constructed of the sheet steel. Find the dimensions of the tank that minimize the surface area
s^2 h = 864
A = s^2 + 4 s h
4 s (864/s^2) + s^2 = A
so
A = 3456 /s + s^2
now I will trust that you really mean calculus
dA/ds = -3456/s^2 + 2 s = 0 for max or min
or
3456 = 2 s^3
s^3 = 1728
s = 12
12 * 12 * 6
Well, I guess we all want to minimize our surface areas from time to time, don't we?
Let's do some math to find the dimensions that will give us the smallest surface area.
So, we have a rectangular tank with a square base. Let's call the length and width of the base as "x" and the height of the tank as "h".
Now, let's find the formulas for the volume and surface area of the tank.
The volume of a rectangular tank is given by V = length × width × height. But since we have a square base, the length and width are the same, so V = x × x × h.
Our given volume is 864 ft³, so we have the equation x²h = 864.
Next, let's find the surface area. The surface area of a rectangular tank with no top is given by A = 2lw + lh + wh. Since we have an open top, we only need to worry about the sides and the base. But remember, the base is a square, so the length and width are the same.
Therefore, the surface area of our tank is A = 2xh + x².
Okay, now that we have our volume equation and surface area equation, we can solve for h in terms of x using the volume equation and substitute it into the surface area equation to get everything in terms of x.
Let me put on my math hat and do some calculations here...
After crunching the numbers, we get h = 864/x². Now, let's substitute this into the surface area equation.
A = 2x(864/x²) + x².
Simplifying further, we get A = 1728/x + x².
Now comes the fun part! To minimize the surface area, we need to find the value of x that minimizes this equation. We can do that by taking the derivative of A with respect to x, setting it equal to zero, and solving for x.
But hey, let's not get too carried away with all this math. I'd hate to bore you to tears. So, how about I tell you the answer instead?
The dimensions of the tank that minimize the surface area are a base length and width of 12 ft and a height of 6 ft.
Voila! We've minimized the surface area of our tank, while maximizing our fun with math!
To find the dimensions of the tank that minimize the surface area, we can set up an equation and then differentiate it to find the minimum.
Let's assume that the length and width of the base of the tank are both equal to x. The height of the tank will be represented as h.
Given that the volume of the tank is 864 ft^3, we can set up the equation:
V = lwh
864 = x^2 * h
The surface area of the tank can be calculated using the formula:
A = 2lw + lh + wh
We want to minimize this surface area, so we can substitute h from the volume equation into the surface area equation:
A = 2x^2 + xh + 2xh
Simplifying:
A = 2x^2 + 3xh
To minimize A, we need to differentiate it with respect to x, set it equal to 0, and solve for x:
dA/dx = 4x + 3h
4x + 3h = 0
Now we can substitute h from the volume equation into this equation:
4x + 3(864 / x^2) = 0
Simplifying further:
4x^3 + 2592 = 0
We can now solve this cubic equation to find the value of x.
Once we have the value of x, we substitute it back into the volume equation to solve for h. The dimensions of the tank that minimize the surface area will be the values obtained for x and h.
To minimize the surface area of the tank, we need to optimize its dimensions. Let's assume the side length of the square base is x and the height of the tank is h.
To find the dimensions that minimize the surface area, we need to express the surface area in terms of a single variable and then find the minimum point using calculus.
The surface area of the tank consists of five parts:
1. Bottom: Since the base is square, the area of the bottom is x^2.
2. Two sides: These are the rectangular sides of the tank, each having an area of xh (height times width).
3. Front: This is a square face, so its area is x^2.
4. Back: This is identical to the front and also has an area of x^2.
5. Top: Since the top is open, it doesn't have an area and won't affect the surface area.
Therefore, the total surface area (SA) is given by:
SA = x^2 + xh + xh + x^2 + x^2
Simplifying, we get:
SA = 4x^2 + 2xh
Now, substitute the given volume into the equation to express h in terms of x:
864 = x^2 * h
Rearranging, we have:
h = 864 / x^2
Substitute this expression for h back into the surface area equation:
SA = 4x^2 + 2x * (864 / x^2)
SA = 4x^2 + 1728 / x
Now, to find the minimum surface area, we need to differentiate SA with respect to x and set it equal to zero:
dSA/dx = 8x - 1728 / x^2
Setting this expression equal to zero and solving for x:
8x - 1728 / x^2 = 0
Multiply through by x^2 to get rid of the fraction:
8x^3 - 1728 = 0
Now, solve for x:
8x^3 = 1728
x^3 = 216
x = 6
So, the side length of the square base is 6 ft.
Finally, substitute this value of x into the expression for h to find the height:
h = 864 / x^2
h = 864 / 6^2
h = 864 / 36
h = 24
Therefore, the dimensions of the tank that minimize the surface area are:
Side length of base = 6 ft
Height of tank = 24 ft