If $2000 m^{2}$ of material is used to to construct a box......,then what is the largest possible volume of the box?

Question

If $2000 m^{2}$ of material is used to to construct a rectangular box with a square base and an open top,then what is the largest possible volume of the box?

I put an equation for the volume :

$V = x(y-2x)^2$ .. with $y^2=2000$ (for the area)

then I took the derivative (to find the Max.) and made it equal to $0$ but then the equation couldn't be solved for real solutions

Lost !

Answer 1

If you set the base length to be $x$, and the height of the box to be $y$, then the volume of the box is $$V=x^2y$$ The box is open, so the constraint function is $$x^2+4xy=2000$$ $$y=\frac{2000-x^2}{4x}$$ Substitute $y$ into $V=x^2y$, we can solve for maximum $$V=x^2 \times \frac{2000-x^2}{4x}$$ $$V=\frac{1}{4}\times(2000x-x^3)$$ We can solve for $$x=20\sqrt{\frac{5}{3}} \approx 25.8199$$

Answer 2

The question states you have a box with a square base. So lets call the side length of this square base $x$ and lets call the height of the box $h$.

We can then say that the volume of the box is given by:$$V=x^2h\tag{1}$$and the surface area of the box will be that of the base plus the four sides, this gives:$$S=x^2+4xh=2000\tag{2}$$Use (2) to express $h$ in terms of $x$

Then substitute this into (1) and use calculus to solve the rest.

Answer 3

Hint: Let $x$ be the base of the square, and $y$ be the height of the box. Then the volume is given by $V(x,y)= x^2y$. But, you know that $A(x,y) = x^2 + 4xy = 2000$. You can isolate $y$ , substitute it in the expression for $V(x,y)$ and solve it like a single variable calculus problem, or you can use Lagrange Multipliers: $$\begin{cases} \nabla V(x,y) = \lambda \nabla A(x,y), \\ A(x,y) = 2000.\end{cases} \quad \lambda \in \Bbb R.$$

The first condition can also be expressed by: $$\begin{vmatrix} \frac{\partial V}{\partial x} & \frac{\partial V}{\partial y} \\ \frac{\partial A}{\partial x} & \frac{\partial A}{\partial y}\end{vmatrix} = 0.$$

Answer 4

If we assume your formula for $V$ makes any sense:

$y^2=2000$ gives $y=\sqrt{2000}$, lets call that $s$. Plugging that into your formula gives: $$ V=x(s-2x)^2=x(2000+4x^2-4sx)=4x^3-4sx^2+2000s $$ Differentiating gives: $$ V'=12x^2-8sx $$ Setting that equal to $0$, and reducing we get: $12x^2=8sx$, $12x=8s$, $x=\frac{2}{3}\sqrt{2000}$, so what ever $x$ is supposed to be there are real solutions, so it seems you made a mistake when you tried to do it.