Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - Review - Exercises - Page 1024: 64

Answer

Largest volume of a box is $=11664 in^3$ or, 115664 cubic inches with dimensions: $x=18,y=36$. or Largest volume of a box is $=11664 in^3$ or, 115664 cubic inches with dimensions: $36 \times 18 \times 18$.

Work Step by Step

Need to apply Lagrange Multipliers Method to determine the dimensions of a rectangular box of maximum volume. we have $\nabla f=\lambda \nabla g$ Formula to calculate the maximum dimension is $4x+y=108$ Formula to find the volume of a box is $V=x^2y$ Simplify to get the value of $x,y$ and $z$ $\dfrac{dV}{dx}=12x(18-x)$ This gives, $x=0, y=108$; $V(0)=0$ $x=18, y=36$; $V(18)=11664$ $x=27, y=0$; $V(27)=0$ Apply Second derivative test: Some noteworthy points to calculate the local minimum, local maximum and saddle point of $f$. 1. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum. 2.If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum. 3. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is a not a local minimum and local maximum or, a saddle point. Thus, critical points are: $(18,36)$ Therefore, the Largest volume of a box is $=11664 in^3$ or, 115664 cubic inches with dimensions: $x=18,y=36$. or Largest volume of a box is $=11664 in^3$ or, 115664 cubic inches with dimensions: $36 \times 18 \times 18$.
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