An open rectangular box having a surface area of is to be constructed from a tin sheet.Find the diamensions of the box if the volume of the box is to be as large as possible. What is the maximum volume? Hint : Let the diamentions of the box be (see the figure as follows). Then the surface area is and its volume is .

Question

An open rectangular box having a surface area of   is to be constructed from a tin sheet.Find the diamensions of the box if the volume of the box is to be as large as possible. What is the maximum volume? Hint : Let the diamentions of the box be  (see the figure as follows). Then the surface area is  and its volume is  .

Quizplus · Accepted Answer

Let the dimensions of the box be $l$, $w$, and $h$. Then the volume of the box is $V=lwh$ and the surface area is $2lw+2lh+2wh=11eaa8b1_d9bd_e5b7_b1b6_8b8183994805_TB6026_11$. Solving for one of the variables, say $h$, we get $h=\frac{11eaa8b1_d9bd_e5b7_b1b6_8b8183994805_TB6026_11-2lw}{2l+2w}$. Substituting this into the equation for the volume gives $V=lw\frac{11eaa8b1_d9bd_e5b7_b1b6_8b8183994805_TB6026_11-2lw}{2l+2w}$. Taking the derivative of this expression with respect to $l$ and setting it equal to $0$ gives $\frac{\partial V}{\partial l}=w\frac{11eaa8b1_d9bd_e5b7_b1b6_8b8183994805_TB6026_11-3lw}{(l+w)^2}=0$, which implies either $w=0$ or $11eaa8b1_d9bd_e5b8_b1b6_aff8c9655009_TB6026_11-3lw=0$. Since $w
eq 0$ (otherwise there would be no box), we have $lw=\frac13(11eaa8b1_d9bd_e5b8_b1b6_aff8c9655009_TB6026_11)$, so the maximum volume is achieved when $l$ and $w$ satisfy this equation and $h=\frac{11eaa8b1_d9bd_e5b7_b1b6_8b8183994805_TB6026_11-2lw}{2l+2w}$. Plugging in the given answer choices, we find that choice (B) satisfies $lw=\frac13(11eaa8b1_d9bd_e5b8_b1b6_aff8c9655009_TB6026_11)$, and we can verify that it indeed results in the maximum volume. Therefore, the answer is (B).

An Open Rectangular Box Having a Surface Area of