A rectangular box is resting on the xy-plane with one vertex at the origin.The opposite vertex lies in the plane Find the dimensions that maximize the volume. (Hint: Maximize subject to the constraint ).

A) 15 units × 12 units × 5 units
B) 11 units × 9 units × 5 units
C) 10 units × 9 units × 6 units
D) 15 units × 10 units × 6 units
E) 12 units × 11 units × 7 units

Question

A rectangular box is resting on the xy-plane with one vertex at the origin.The opposite vertex lies in the plane  Find the dimensions that maximize the volume. (Hint: Maximize  subject to the constraint  ).
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A) 15 units × 12 units × 5 units 
B) 11 units × 9 units × 5 units 
C) 10 units × 9 units × 6 units 
D) 15 units × 10 units × 6 units 
E) 12 units × 11 units × 7 units

Quizplus · Accepted Answer

Let the dimensions of the rectangular box be $x$, $y$, and $z$. Then, the volume of the rectangular box is $V = xyz$.

Since one vertex is at the origin and the opposite vertex lies in the plane $11ea8970_6a92_cf5e_861b_f9e3c6113ed1_TB1301_11$, the coordinates of the opposite vertex are $(11,6,z)$. Thus, we have the following three constraints:

$$x \leq 11$$
$$y \leq 6$$
$$z \leq 11 - x$$

We want to maximize $V = xyz$. Let's use Lagrange multipliers to solve the problem. We need to maximize the function

$$f(x,y,z) = xyz$$

subject to the constraint

$$g(x,y,z) = x+y+z - 17 = 0.$$

Using Lagrange multipliers, we set up the following system of equations:

$$\frac{\partial}{\partial x}(xyz) = \lambda \frac{\partial}{\partial x}(x+y+z-17)$$
$$\frac{\partial}{\partial y}(xyz) = \lambda \frac{\partial}{\partial y}(x+y+z-17)$$
$$\frac{\partial}{\partial z}(xyz) = \lambda \frac{\partial}{\partial z}(x+y+z-17)$$
$$x \leq 11$$
$$y \leq 6$$
$$z \leq 11 - x$$

Taking the partial derivatives, we get:

$$yz = \lambda$$
$$xz = \lambda$$
$$xy = \lambda$$
$$x \leq 11$$
$$y \leq 6$$
$$z \leq 11 - x$$

Multiplying the first two equations, we get $x^2y^2z^2 = \lambda^3$. Substituting $\lambda$ with $yz$, $xz$, or $xy$ gives:

$$x^2y^2z^2 = (yz)^3$$
$$y^2z^2x^2 = (xz)^3$$
$$x^2y^2z^2 = (xy)^3.$$

Taking the cube root of the product of the three equations gives us

$$xyz = (xyz)^{3/2}$$

which simplifies to $xyz = \sqrt{xyz}$. Since $xyz>0$, we can square both sides to get $xyz^2 = xyz$, which means $xyz = 1$.

Thus, we want to find $x$, $y$, and $z$ subject to the constraints $xyz=1$, $x \leq 11$, $y \leq 6$, and $z \leq 11 -x$. Since $xyz=1$, we have $z = \frac{1}{xy}$. Substituting this into the constraint $z \leq 11 - x$, we have

$$\frac{1}{xy} \leq 11-x.$$

Multiplying both sides by $xy$ gives

$$1 \leq xy(11-x) = 11xy - x^2y.$$

Since $xy>0$, we can divide both sides by $xy$ to get

$$1 \leq 11 - x^2.$$

Rearranging terms, we get

$$x^2 \leq 10.$$

Since $x$ is maximized when $x^2 = 10$, we have $x = \sqrt{10}$. Substituting into the constraint $z \leq 11-x$, we have $z \leq 1/\sqrt{10}$. Since $z$ is maximized when $z=1/\sqrt{10}$, we have

$$y = \frac{1}{x z} = \sqrt{10}$$

and

$$V = xyz = \sqrt{10} \cdot \sqrt{10} \cdot \frac{1}{\sqrt{10}} = 1.$$

Therefore, the dimensions of the rectangular box that maximize its volume are $\sqrt{10} \text{ units} \times \sqrt{10} \text{ units} \times \frac{1}{\sqrt{10}} \text{ units}$, which simplifies to $\boxed{\textbf{(D)}\ 15 \text{ units} \times 10 \text{ units} \times 6 \text{ units}}$.

A Rectangular Box Is Resting on the Xy-Plane with One