Answer to Question #249311 in Calculus for Anmol

Question #249311

What are dimensions of rectangular prism with the largest possible volume that can be construed using 2400 square inches of surface area ?


1
Expert's answer
2021-10-11T15:58:30-0400

If we are assuming a fully rectangular prism---right angles all around---then


"2xy+2(x+y)z=2400"

"xy+(x+y)z=1200"

"z=\\dfrac{1200-xy}{x+y}"

"V(x, y)=xy(\\dfrac{1200-xy}{x+y})"

"V(x, y)=\\dfrac{1200xy-x^2y^2}{x+y}"

"V_x=\\dfrac{(1200y-2xy^2)(x+y)-(1200xy-x^2y^2)}{(x+y)^2}"

"=\\dfrac{y^2(1200-x^2-2xy)}{(x+y)^2}"

"V_y=\\dfrac{(1200x-2x^2y)(x+y)-(1200xy-x^2y^2)}{(x+y)^2}"

"=\\dfrac{x^2(1200-y^2-2xy)}{(x+y)^2}"

Find the critical (stationary) point(s)

"\\begin{matrix}\n V_x=0 \\\\\n V_y=0\n\\end{matrix}"

"y^2(1200-x^2-2xy)=0"

"x^2(1200-y^2-2xy)=0"

Then for "x>0, y>0"


"x^2-y^2=0"

"(x-y)(x+y)=0"

"x=y"

"1200-x^2-2x^2=0"

"x^2=400, x>0"

"x=20=y"

"z=\\dfrac{1200-20(20)}{20+20}=20"


"V=20(20)(20)=8000 (in^3)"


"V_{xx}=\\dfrac{-2y^2(y^2+1200)}{(x+y)^3}"

"V_{yy}=\\dfrac{-2x^2(x^2+1200)}{(x+y)^3}"

"V_{xy}=\\dfrac{-2xy(x^2+y^2+3xy-1200)}{(x+y)^3}=V_{yx}"

"V_{xx}(20,20)=\\dfrac{-2(20)^2((20)^2+1200)}{(20+20)^3}=-20<0"

"V_{yy}(20,20)=\\dfrac{-2(20)^2((20)^2+1200)}{(20+20)^3}=-20"

"V_{xy}(20,20)=\\dfrac{-2(20)(20)((20)^2+(20)^2+3(20)(20)-1200)}{(20+20)^3}"

"=-10"


"D=\\begin{vmatrix}\n -20 & -10 \\\\\n -10 & -20\n\\end{vmatrix}=300>0"

The volume has a relative maximum at "(20, 20)." Since there is the only stationary point for "x>0, y>0," then the volume has the absolute maximum for "x>0, y>0" at "(20, 20)."

We have the cube with side "20" inches.


"V_{max}=20(20)(20)=8000\\ in^3"


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