A tank with a square base of side x has a capacity of 1000m^3. If the total area is pm^3, show that p=2x^2+4000/x. Hence find the least area of the flat sheet that may be used to build the tank.
Please show workings
#thanks
v = x^2h, so
h = 1000/x^2
a = 2x^2 + 4xh = 2x^2 + 4000/x
da/dx = 4x - 4000/x^2
= 4(x^3-1000)/x^2
so, da/dx = 0 when x=10
a = 2*100 + 4*10*1000/100 = 600
as usual, area is minimum when the shape is a cube.
To solve this problem, let's break it down step-by-step:
1. Find the dimensions of the tank:
Given that the tank has a square base of side x, the volume of the tank is 1000m^3. Since the volume of a cuboid is given by the formula V = L x W x H, where L, W, and H represent the length, width, and height respectively, we can write the equation as:
x^2 * H = 1000
2. Solve for the height:
Solving for H in the equation x^2 * H = 1000, we get:
H = 1000 / x^2
3. Find the total area:
The total area is given by the formula A = 2(LW + LH + WH). Since the base of the tank is a square, we can rewrite the formula as:
A = 2(x^2 + xH + xH)
= 2(x^2 + 2xH)
= 2x^2 + 4xH
4. Substitute the height value:
Plugging in the value of H we found earlier, we get:
A = 2x^2 + 4x(1000 / x^2)
= 2x^2 + 4000 / x
5. Show that p = 2x^2 + 4000 / x:
We are given that the total area is p, so p = A. Therefore, we can conclude that p = 2x^2 + 4000 / x.
6. Find the least area of the flat sheet:
To find the least area that can be used to build the tank, we need to minimize the value of p. To do this, we can find the derivative of p with respect to x and set it equal to zero:
dp/dx = 4x - 4000 / x^2 = 0
Solving this equation, we get:
4x = 4000 / x^2
4x^3 = 4000
x^3 = 1000
x = ∛1000
x = 10
Now that we know the value of x, we can substitute it back into the equation for p:
p = 2x^2 + 4000 / x
= 2(10)^2 + 4000 / 10
= 200 + 400
= 600
Therefore, the least area of the flat sheet that may be used to build the tank is 600m^2.
To solve this problem, we need to use the given information about the tank's capacity and total surface area.
Let's start by finding the volume of the tank. The volume of a rectangular prism (such as this tank) is obtained by multiplying the length, width, and height. Since the base of the tank is square, the length and width are both equal to x, and the height is also x.
Therefore, the volume of the tank is:
Volume = length × width × height = x × x × x = x^3
Given that the capacity of the tank is 1000 m^3, we can set up the following equation:
x^3 = 1000
Now, let's find the total surface area of the tank. The total surface area of a rectangular prism is obtained by adding the areas of all its six faces. In this case, we have four identical square faces (each with an area of x^2) and two identical rectangular faces (each with an area of x × x, which simplifies to x^2).
Therefore, the total surface area of the tank is:
Surface Area = 4 × (area of square face) + 2 × (area of rectangular face)
Surface Area = 4x^2 + 2x^2
Surface Area = 6x^2
Given that the total surface area is p m^2, we can set up the following equation:
6x^2 = p
Now, we have two equations:
x^3 = 1000 (equation 1)
6x^2 = p (equation 2)
To find p in terms of x, we can rearrange equation 2:
p = 6x^2
Now, substitute equation 1 into equation 2:
p = 6(1000/x)^2
p = 6000/x^2
Finally, simplify p in terms of x:
p = 2x^2 + 4000/x
To find the least area of the flat sheet needed to build the tank, we need to find the minimum value of p. To do that, we can differentiate the equation with respect to x, set it equal to zero, and solve for x:
dp/dx = 4x - 4000/x^2 = 0
4x = 4000/x^2
4x^3 = 4000
Now, solve for x:
x^3 = 1000
x = ∛1000
x = 10
Substitute x = 10 into the equation for p:
p = 2(10)^2 + 4000/10
p = 200 + 400
p = 600
Therefore, the least area of the flat sheet needed to build the tank is 600 m^2.