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a rectangular cardboard box is made with a square base and an open top.…

Question

a rectangular cardboard box is made with a square base and an open top. the box has a volume of 4 cubic meters.
a. find the dimensions of the box that uses the least cardboard.
height of the box: m
length of the side of the base: m
b. find the least amount of cardboard needed to construct the box.
answer: m²

Explanation:

Step1: Define variables

Let the side length of the square base be \(x\) meters and the height be \(h\) meters.
The volume \(V=x^{2}h = 4\), so \(h=\frac{4}{x^{2}}\).
The surface area \(S=x^{2}+4xh\) (since the top is open).

Step2: Substitute \(h\) into the surface - area formula

Substitute \(h = \frac{4}{x^{2}}\) into \(S\):
\(S=x^{2}+4x\cdot\frac{4}{x^{2}}=x^{2}+\frac{16}{x}\), \(x>0\).

Step3: Find the derivative of \(S\)

Differentiate \(S\) with respect to \(x\):
\(S^\prime=2x-\frac{16}{x^{2}}\).

Step4: Set the derivative equal to zero

Set \(S^\prime = 0\):
\(2x-\frac{16}{x^{2}}=0\).
Multiply through by \(x^{2}\) to get \(2x^{3}-16 = 0\).
\(x^{3}=8\), so \(x = 2\).

Step5: Find the second - derivative

Differentiate \(S^\prime\) with respect to \(x\):
\(S^{\prime\prime}=2+\frac{32}{x^{3}}\).
When \(x = 2\), \(S^{\prime\prime}(2)=2+\frac{32}{8}=2 + 4=6>0\), so \(S\) has a minimum at \(x = 2\).

Step6: Find the height

When \(x = 2\), \(h=\frac{4}{x^{2}}=\frac{4}{4}=1\).

Step7: Calculate the minimum surface area

Substitute \(x = 2\) into \(S\):
\(S=x^{2}+4xh=2^{2}+4\times2\times1=4 + 8=12\).

Answer:

a. Height of the box: \(1\) m, Length of the side of the base: \(2\) m.
b. \(12\) \(m^{2}\).