A soup company is constructing an open-top, square-based, rectangular metal tank that will have a volume of 32 ft^3. What dimensions of the tank will minimize the surface area? What is the minimum surface area?
Guide On Rating System
Vote
Let's assume the length of the rectangular tank is L ft, the width is W ft, and the height is H ft.
Since the tank is square-based, the length and width both equal L ft.
The volume of a rectangular tank is given by the formula: V = L * W * H
Given that the volume is 32 ft^3, we can write the equation as:
32 = L * L * H
32 = L^2 * H ----(Equation 1)
Now, let's find the formula for surface area.
The surface area of the rectangular tank consists of three pairs of faces: (L * H), (L * W), and (W * H).
Surface Area (A) = 2(L * H) + 2(L * W) + 2(W * H)
A = 2(L * H + L * W + W * H)
A = 2(LH + LW + WH) ----(Equation 2)
To minimize the surface area, we need to find the dimensions that will minimize the expression in Equation 2.
Now, let's substitute the value of H from Equation 1 into Equation 2.
A = 2(LH + LW + WH)
A = 2(L^2 * H + L * W + W * H) ----(Equation 3)
Substituting 32/L^3 for H (from Equation 1), we can rewrite Equation 3 as:
A = 2(L^2 * (32 / L^3) + L * W + W * (32 / L^3))
A = 2(32 / L + LW + 32W / L^3) ----(Equation 4)
To minimize A, we can take the derivative of A with respect to L and set it equal to zero.
dA/dL = 0
Differentiating Equation 4 with respect to L:
dA/dL = 2(-32 / L^2 + W - 96W / L^4) = 0
Simplifying the equation, we have:
-32 / L^2 + W - 96W / L^4 = 0
-32L^2 + WL^4 - 96W = 0
-32L^2 + WL^2(L^2-96) = 0
W(L^2 - 96) = 32L^2
W = 32L^2 / (L^2 - 96)
Now, let's find the value of L that satisfies the minimum surface area. We can substitute the value of W in terms of L into Equation 2.
A = 2(LH + LW + WH)
A = 2(L * (32 / L^3) + L * (32L^2 / (L^2 - 96) + (32L^2 / (L^2 - 96)) * (32 / L^3)))
Simplifying the equation, we get:
A = 2(32L^3 / L^3 + 32L^3 + 32L^3)
A = 2(96L^3) = 192L^3 ----(Equation 5)
To find the minimum surface area (A), we need to find the value of L that will minimize Equation 5. We can take the derivative of Equation 5 with respect to L.
dA/dL = 0
Differentiating Equation 5 with respect to L:
dA/dL = 0 = 576L^2
L = 0 or L is undefined
Since L cannot be zero for a tank, L is undefined.
This means that the surface area cannot be minimized for a rectangular tank with a volume of 32 ft^3 and square base.
Got a question with my answer?
Cheapmath.com is completely free!
Asking question is free.
Getting answers is free.
+99
Vote
Great work Koala AI.
Got a question with my answer?
Cheapmath.com is completely free!
Asking question is free.
Getting answers is free.
+99