Minimizing surface area of box with given volume

[Anthonyk2013]

Q4 here.

4. A lidless box with square ends is to be made from a thin sheet of metal. Determine the least area of the metal for which the volume of the box is 3.5m³.

Wrong answer but I'm not sure where I've gone wrong:

\(\displaystyle \mbox{Volume: }\, V\, =\, 3.5m^3\)

\(\displaystyle \mbox{Volume of box: }\, x^2 y\)

\(\displaystyle \mbox{Area of base: }\, x^2\)

\(\displaystyle \mbox{Area of four sides: }\, 4xy\)

\(\displaystyle \mbox{Material needed: }\, x^2\, +\, 4xy\)

\(\displaystyle x^2 y\, =\, 3.5\)

\(\displaystyle y\, =\, \dfrac{3.5}{x^2}\)

\(\displaystyle 4xy\, =\, 4x\, \cdot\, \dfrac{3.5}{x^2}\, =\, \dfrac{14}{x}\)

\(\displaystyle M\, =\, x^2\, +\, \dfrac{14}{x}\)

\(\displaystyle \dfrac{dM}{dx}\, =\, 2x\, +\, \left(-\dfrac{14}{x^2}\right)\, =\, 2x\, -\, \dfrac{14}{x^2}\)

\(\displaystyle \cdot\, 2x\, -\, \dfrac{14}{x^2}\, =\, 0\)

\(\displaystyle 2x^3\, -\, 14\, =\, 0\)

\(\displaystyle x\, =\, \sqrt[3]{\strut 7\,}\, \approx\, 1.912\)

\(\displaystyle y\, =\, \dfrac{3.5}{x^2}\, \approx\, \dfrac{3.5}{1.912^2}\, \approx\, 0.957\)

\(\displaystyle A\, =\, 4\, (\, 1.912\,)\, (\, 0.957\,)\, =\, 7.31\)

 

[daon2]

Square ends, in my mind, does not mean square base. I would say A= 2x^2+3xy

 

[Anthonyk2013]

\(\displaystyle V\, =\, x^2\, y\, =\, 3.5m^3\)

\(\displaystyle A\, =\, 2x^2\, +\, 3xy\)

\(\displaystyle \mbox{Material }\, M\, \mbox{ needed: }\, M\, =\, x^2\, +\, 4xy\)

\(\displaystyle y\, =\, \dfrac{3.5}{x^2}\)

\(\displaystyle A\, =\, 2x^2\, +\, 3x\, \left(\dfrac{3.5}{x^2}\right)\)

\(\displaystyle A\, =\, 2x^2\, +\, \dfrac{10.5}{x}\)

\(\displaystyle \dfrac{dA}{dx}\, =\, 4x\, -\, \dfrac{10.5}{x^2}\)

\(\displaystyle 4x\, -\, \dfrac{10.5}{x^2}\, =\, 0\)

\(\displaystyle 4x\, =\, \dfrac{10.5}{x^2}\)

\(\displaystyle x^3\, =\, \dfrac{10.5}{4}\)

\(\displaystyle x\, =\, \sqrt[3]{\strut 2\, \dfrac{5}{8}\,}\)

\(\displaystyle x\, \approx\, 1.3794\)

\(\displaystyle A\, =\, 2\, \bigg(\, 1.3794^2\, \bigg)\, +\, 3\, \bigg(\, 1.3794\, \bigg)\left(\,\dfrac{3.5}{1.3794^2}\,\right)\, =\, 11.51m^2\)

Is this solution correct?

 

[lookagain]

Q4 here.

4. A lidless box with square ends is to be made from a thin sheet of metal.
Determine the least area of the metal for which the volume of the box is 3.5m³.

Wrong answer

but I'm not sure where I've gone wrong: \(\displaystyle \ \ \ \ \)You forgot to add x^2 as part of A.

\(\displaystyle \mbox{Volume: }\, V\, =\, 3.5m^3\)

\(\displaystyle \mbox{Volume of box: }\, x^2 y\)

\(\displaystyle \mbox{Area of base: }\, x^2\)

\(\displaystyle \mbox{Area of four sides: }\, 4xy\)

\(\displaystyle \mbox{Material needed: }\, x^2\, +\, 4xy\)

\(\displaystyle x^2 y\, =\, 3.5\)

\(\displaystyle y\, =\, \dfrac{3.5}{x^2}\)

\(\displaystyle 4xy\, =\, 4x\, \cdot\, \dfrac{3.5}{x^2}\, =\, \dfrac{14}{x}\)

\(\displaystyle M\, =\, x^2\, +\, \dfrac{14}{x}\)

\(\displaystyle \dfrac{dM}{dx}\, =\, 2x\, +\, \left(-\dfrac{14}{x^2}\right)\, =\, 2x\, -\, \dfrac{14}{x^2}\)

\(\displaystyle \cdot\, 2x\, -\, \dfrac{14}{x^2}\, =\, 0\)

\(\displaystyle 2x^3\, -\, 14\, =\, 0\)

\(\displaystyle x\, =\, \sqrt[3]{\strut 7\,}\, \approx\, 1.912 \ \ \ \ \) <----- It is closer to 1.913.

\(\displaystyle y\, =\, \dfrac{3.5}{x^2}\, \approx\, \dfrac{3.5}{1.912^2}\, \approx\, 0.957 \ \ \ \ \)

\(\displaystyle A\, =\, 4\, (\, 1.912\,)\, (\, 0.957\,)\, =\, 7.31\)

Suppose "square ends" do mean "square bases." (I see the other perspective as was pointed out.)

A = x^2 + 4xy

A = x^2 + (4x)[3.5/(x^2)]

A = x^2 + 14/x

x = cube root of 7 ~ 1.913

A ~ 10.98


Anthonyk2013, is 10.98 square meters the answer given?