Question

How do you 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 $x+5y+8z=40$?

Answer 1

In this problem, we have to find the volume of the largest rectangular box whose plane is given. We can find the first derivative to find the critical point and we can find what type of critical point is that, to find the largest rectangular box. We can then find the second derivative to prove for the condition of the local maximum at which the box will be largest.

Complete step by step solution:
We know that the plane equation is $x+5y+8z=40$ which is one of the rectangular boxes.
We can now assume a point $\left( a,b,c \right)$ in the plane $x+5y+8z=40$.
We have to find the volume of the box.
We know that the volume of the box is, $V=a\times b\times c$.
We know that, since the point is on the plane $x+5y+8z=40$ then its satisfies its equation as,
$\Rightarrow a+5b+8c=40$
We can write it as,
$\Rightarrow a=-5b-8c+40$
Now we can substitute this in volume formula, we get

& \Rightarrow V=\left( -5b-8c+40 \right)bc \\\ & \Rightarrow V=-5{{b}^{2}}c-8b{{c}^{2}}+40bc \\\ \end{aligned}$$ We can assume, let $$f\left( x,y \right)=-5{{x}^{2}}y-8x{{y}^{2}}+40xy$$ for $$x,y > 0,-5b-8c+40 > 0$$. Since we need to find the volume of largest rectangular box, we have $$V=\max \left\\{ f\left( x,y \right) \right\\}$$ Here we have to find the maximum of $$f$$. We can now find the critical points by equation partial derivatives with 0, we get $$\begin{aligned} & \Rightarrow {{f}_{x}}\left( x,y \right)=-10xy-8{{y}^{2}}+40y=0...\left( 1 \right) \\\ & \Rightarrow {{f}_{y}}\left( x,y \right)=-5{{x}^{2}}-16xy+40x=0...\left( 2 \right) \\\ \end{aligned}$$ We can now divide the equation (1) by y and equation (2) by x, we get $$\begin{aligned} & \Rightarrow -10x-8y+40...\left( 3 \right) \\\ & \Rightarrow -5x-16y+40...\left( 4 \right) \\\ \end{aligned}$$ We can now solve the above equations by dividing 2 in the equation (3) and subtracting it from the equation (4). $$\begin{aligned} & \Rightarrow -5x-4y+20+5x+16y-40=0 \\\ & \Rightarrow 12y=20 \\\ & \Rightarrow y=\dfrac{5}{3} \\\ \end{aligned}$$ We can now find x by substituting y, we get $$\begin{aligned} & \Rightarrow -10x-8\left( \dfrac{5}{3} \right)+40=0 \\\ & \Rightarrow -10x-\dfrac{40}{3}+40=0 \\\ & \Rightarrow x=\dfrac{8}{3} \\\ \end{aligned}$$ Hence $$\left( \dfrac{8}{3},\dfrac{5}{3} \right)$$ is the critical point. We can now find the second derivative to find which type of critical point. We can now find the partial derivative of $$f$$, let $$D=D\left( a,b \right)={{f}_{xx}}\left( a,b \right){{f}_{yy}}\left( a,b \right)-{{\left[ {{f}_{xy}}\left( a,b \right) \right]}^{2}}$$ …… (5) If $$D > 0\text{ and }{{f}_{xx}}\left( a,b \right) > 0,\text{ then }f\left( a,b \right)$$ is a local minimum. If $$D > 0\text{ and }{{f}_{xx}}\left( a,b \right)<0,\text{ then }f\left( a,b \right)$$is a local maximum. If $$D<0,f\left( a,b \right)$$ is a saddle point. We can now find the second derivative, we get $$\begin{aligned} & \Rightarrow {{f}_{xx}}\left( x,y \right)=-10y \\\ & \Rightarrow {{f}_{yy}}\left( x,y \right)=-16x \\\ & \Rightarrow {{f}_{xy}}\left( x,y \right)=-10x-16y+40 \\\ \end{aligned}$$ We can substitute the above values in (5), we get $$\begin{aligned} & \Rightarrow D\left( \dfrac{8}{3},\dfrac{5}{3} \right)={{f}_{xx}}\left( \dfrac{8}{3},\dfrac{5}{3} \right){{f}_{yy}}\left( \dfrac{8}{3},\dfrac{5}{3} \right)-{{\left[ {{f}_{xy}}\left( \dfrac{8}{3},\dfrac{5}{3} \right) \right]}^{2}} \\\ & \Rightarrow D\left( \dfrac{8}{3},\dfrac{5}{3} \right)=\left( -\dfrac{50}{3}\times -\dfrac{128}{3} \right)-\left( -\dfrac{160}{3}+40 \right) \\\ & \Rightarrow D\left( \dfrac{8}{3},\dfrac{5}{3} \right)=\dfrac{680}{3} > 0 \\\ & \Rightarrow {{f}_{xx}}\left( \dfrac{8}{3},\dfrac{5}{3} \right)=-\dfrac{50}{3}<0 \\\ \end{aligned}$$ Therefore, $$\left( \dfrac{8}{3},\dfrac{5}{3} \right)$$ is a local maxima. We can now find the volume, $$\begin{aligned} & \Rightarrow V=f\left( \dfrac{8}{3},\dfrac{5}{3} \right)=\left( -5{{\left( \dfrac{8}{3} \right)}^{2}}\dfrac{5}{3} \right)-\left( 8\left( \dfrac{8}{3} \right){{\left( \dfrac{5}{3} \right)}^{2}} \right)+40\left( \dfrac{8}{3} \right)\dfrac{5}{3} \\\ & \Rightarrow V=\dfrac{1600}{27} \\\ \end{aligned}$$ **Therefore, the volume of the largest rectangular box is $$\dfrac{1600}{27}$$ cubic units.** **Note:** Students make mistakes while remembering the conditions for the local maximum and local minimum which are, If $$D > 0\text{ and }{{f}_{xx}}\left( a,b \right) > 0,\text{ then }f\left( a,b \right)$$ is a local minimum. If $$D > 0\text{ and }{{f}_{xx}}\left( a,b \right)<0,\text{ then }f\left( a,b \right)$$ is a local maximum. If $$D<0,f\left( a,b \right)$$ is a saddle point. We should always mention a cubic unit or simply cubic unit whenever we write an answer for the volume.