The dimensions (in m) maximize its volume is mathematically given as
[tex]x=\frac{4\sqrt{3}}{3}[/tex] m for Lenght and Width
[tex]h=\frac{2\sqrt{3}}{3}[/tex] m for height
Generally, the equation for is mathematically given as
The equation for the area is given as
[tex]A=x^2+4xh\\\\A=16=4xh\\\\ A=16-x^2[/tex]
[tex]h=\frac{16-x^2}{4x^2}[/tex]
Volume of box
[tex]V=x^2 *h\\\\V=x^2*(\frac{16-x^2}{4x^2})\\\\V=4x-x^3[/tex]
Therefore, by differentiating V and making x the subject of the formula we have
[tex]V'=4-\frac{3x^2}{4}[/tex]
Therefore, with V' =0
[tex]x^2=16/3[/tex]
[tex]x^2=\sqrt{16/3}\\\\x=\frac{4}{\sqrt{3}}\\\\[/tex]
[tex]x=\frac{4\sqrt{3}}{3}[/tex]
Now we will sub the value of x to the h equation
[tex]\begin{aligned}&h=\frac{16-(\frac{4}{\sqrt{3}})^2}{4 (\frac{4}{\sqrt{3}})} \\\\&h=\frac{16-\frac{16}{3}}{\frac{16}{\sqrt{2}}} \\\\&h=\frac{2 \sqrt{3}}{3} \mathrm{~m}\end{aligned}[/tex]
Now we differntite the derivate of V (V')
[tex]V''=\frac{-3x}{2}\\V''=\frac{3(\frac{4\sqrt{3}}{3})}{2} \\V''=-2\sqrt{3} < 0 \\\\[/tex] at maximium
In conclusion, the dimensions (in m) maximize its volume.
[tex]x=\frac{4\sqrt{3}}{3}[/tex] for Lenght and Width
[tex]h=\frac{2\sqrt{3}}{3}[/tex] for height
Read more about volume
https://brainly.com/question/1578538
#SPJ4
