Question

Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane\(x + 2y + 3z = 6\).

Short answer

The volume of largest rectangular box is\(\frac{4}{3}\)unit cube.

Step-by-step solution

Second derivative test

Suppose the second partial derivatives of\(f\)are continuous on a disk with center\({\bf{(a,b)}}\), and suppose that\({f_x}(a,b) = 0\)and\({f_y}(a,b) = 0\). Let

\(D = D(a,b) = {f_{xx}}(a,b){f_{yy}}(a,b) - {\left( {{f_{xy}}(a,b)} \right)^2}\)

(a) If\(D > 0\)and\({f_{xx}}(a,b) > 0\), then\(f(a,b)\)is a local minimum.

(b) If\(D > 0\)and\({f_{xx}}(a,b) < 0\), then\(f(a,b)\)is a local maximum.

(c) If\(D < 0\), then\(f(a,b)\)is not a local maximum or minimum.

Assumption

Consider a rectangular box in the first octant with three faces in the coordinate planes and one vertex of the rectangular box is in the plane\(x + 2y + 3z = 6\).

Here, \(x,y,z \ge 0\)because, the rectangular box is lies in the first octant.

Let\((x,y,z)\)be the vertex on the plane\(x + 2y + 3z = 6\), and\(x,y\), and\(z\) are the length, width, and height of the box.

Form equation

The volume of a rectangular solid is\(V = \left( {{\rm{ }}length} \right)\left( {width} \right)\left( {height} \right)\).

\(V = xyz\)

The plane equation \(x + 2y + 3z = 6\)for\(z\)as follows:

\(\begin{array}{c}x + 2y + 3z = 6\\3z = 6 - x - 2y\\z = \frac{{6 - x - 2y}}{3}\end{array}\)

Substitute\(z = \frac{{6 - x - 2y}}{3}\)in\(V = xyz\).

\(V(x,y) = \frac{1}{3}\left( {6xy - {x^2}y - 2x{y^2}} \right)\)

Compute partial derivative

Differentiate\(V(x,y)\)with respect to\(x\)by assuming\(y\)as constant:

\(\begin{array}{c}{V_x}(x,y) = \frac{\partial }{{\partial x}}\left( {\frac{1}{3}\left( {6xy - {x^2}y - 2x{y^2}} \right)} \right)\\ = \frac{1}{3}\left( {\frac{\partial }{{\partial x}}(6xy) + \frac{\partial }{{\partial x}}\left( { - {x^2}y} \right) + \frac{\partial }{{\partial x}}\left( { - 2x{y^2}} \right)} \right){\rm{ Use the formula }}\frac{d}{{dx}}{x^n} = n{x^{n - 1}}\\ = \frac{1}{3}\left( {6y\frac{\partial }{{\partial x}}(x) - y\frac{\partial }{{\partial x}}\left( {{x^2}} \right) - 2{y^2}\frac{\partial }{{\partial x}}(x)} \right)\\{V_x}(x,y) = \frac{1}{3}\left( {6y - 2xy - 2{y^2}} \right)\,\,\,\,\,\, - - - (1)\end{array}\)

Similarly \({V_y}(x,y) = \frac{1}{3}\left( {6x - {x^2} - 4xy} \right)\,\,\,\,\, - - - (2)\)

Find critical points

Set the equation (1) equal to zero as follows:

\(\begin{array}{l}\frac{1}{3}\left( {6y - 2xy - 2{y^2}} \right) = 0\\6y - 2xy - 2{y^2} = 0\\2y(3 - x - y) = 0\\y = 0\;or\,\,3 = x + y\end{array}\)

Set the equation (2) equal to zero as follows:

\(\begin{array}{l}\frac{1}{3}\left( {6x - {x^2} - 4xy} \right) = 0\\6x - {x^2} - 4xy = 0\\x(6 - x - 4y) = 0\\x = 0\;\;\;{\rm{ and }}\;\;\;6 = x + 4y\end{array}\)

Solving we get critical points\((0,0),(0,3),(6,0),{\rm{ and }}(2,1)\)

 Step 6: Compute second order partial derivative

To find\(D(x,y)\), find \({V_{xx}}(x,y),{V_{yy}}(x,y)\)and\({V_{xy}}(x,y)\)\(\begin{array}{c}{V_{xx}}(x,y) = \frac{\partial }{{\partial x}}\left( {\frac{1}{3}\left( {6y - 2xy - 2{y^2}} \right)} \right)\\ = \frac{1}{3}\left( {6y\frac{\partial }{{\partial x}}(1) - 2y\frac{\partial }{{\partial x}}(x) - 2{y^2}\frac{\partial }{{\partial x}}(1)} \right)\\{V_{xx}}(x,y) = - \frac{2}{3}y\end{array}\)

And

\(\begin{array}{c}{V_{yy}}(x,y) = \frac{\partial }{{\partial y}}\left( {\frac{1}{3}\left( {6x - {x^2} - 4xy} \right)} \right)\\ = \frac{1}{3}\left( {6x\frac{\partial }{{\partial y}}(1) - {x^2}\frac{\partial }{{\partial y}}(1) - 4x\frac{\partial }{{\partial y}}(y)} \right)\\ = - \frac{4}{3}x\end{array}\)

And

\(\begin{array}{c}{V_{xy}}(x,y) = \frac{\partial }{{\partial x}}\left( {\frac{1}{3}\left( {6x - {x^2} - 4xy} \right)} \right)\\ = \frac{1}{3}\left( {6\frac{\partial }{{\partial x}}(x) - \frac{\partial }{{\partial x}}\left( {{x^2}} \right) - 4y\frac{\partial }{{\partial x}}(x)} \right)\\ = - \frac{{2x + 4y - 6}}{3}\end{array}\)

Compute discriminant

Substitute\({V_{xx}},{V_{yy}}\)and\({V_{xy}}\)in\(D\).

\(\begin{array}{c}D = {V_{xx}}(x,y){V_{yy}}(x,y) - {\left( {{V_{xy}}(x,y)} \right)^2}\\ = \left( { - \frac{2}{3}y} \right) - \frac{4}{3}x - {\left( { - \frac{{2x + 4y - 6}}{3}} \right)^2}\\ = \frac{{ - 4{x^2} - 8xy - 16{y^2} + 24x + 48y - 36}}{9}\end{array}\)

Determine nature of points

At point\((0,3)\):

\(\begin{array}{l}V(0,3) = 0\\{V_{xx}}(0,3) = - 2 < 0\\D( - 0,3) = - 4 < 0\end{array}\)

So neither local maximum nor minimum.

At point\((6,0)\):

\(\begin{array}{c}V(6,0) = 0\\{V_{xx}}(6,0) = 0\\D(6,0) = - 4 < 0\end{array}\)

So neither local maximum nor minimum.

At point \((2,1)\):

\(\begin{array}{c}V(2,1) = \frac{4}{3}\\{V_{xx}}(2,1) = \frac{{ - 2}}{3} < 0\\D(2,1) = \frac{4}{3} > 0\end{array}\)

So local maximum.

Thus, volume of the largest rectangular box in the first octant is\(\frac{4}{3}\).