Printed Page 9999
In Problems 1–6, find the gradient and directional derivative of each function at the indicated point in the direction of a.
f(x,y)=ln(sec(x2+y2)) at (√π8,√π8);a=i+j
∇f(√π8,√π8)=√π2i+√π2j; Duf(√π8,√π8)=√π
f(x,y)=x2−y2x2+y2 at (1,1);a=i−j
f(x,y)=ysin−1x at (0,1); a=3i+j
∇f(0,1)=i; Duf(0,1)=3√1010
f(x,y,z)=ln(xyz) at (1,1,3); a=i+j+3k
f(x,y,z)=ye−x(x2+y2+z2+1) at (0,0,0); a=2i+j+2k
∇f(0,0,0)=j; Duf(0,0,0)=13
f(x,y,z)=(2x+y+z)2+xyz at (1,1,1); a=−i−j−k
In Problems 7–9:
f(x,y)=sec(x2+y2) at (√π8,√π8)
f(x,y)=\dfrac{x^{2}-y^{2}}{x^{2}+y^{2}} at (1,1)
f(x,y)=y\sin ^{-1}x at (0,1)
In Problems 10–12:
x^{2}-y^{2}+z^{2}=4 at (-1,1,2)
2x^{2}+y^{2}=z at (1,0,2)
f(x,y)=y\sin ^{-1}x at ( 0,1)
In Problems 13–15, find the critical points of each function.
z=f(x,y)=x^{2}+xy+y^{2}+6x
The critical point is (-4, 2).
z= f(x,y)=x^{3}-y^{3}+3xy
z=f( x,y) =xy
The critical point is (0, 0).
In Problems 16–18, use the Second Partial Derivative Test to find any local maxima, local minima, and saddle points for each function.
z=f( x,y) =xe^{xy}
z=f( x,y) =\sin x+\sin y
For integers m and n, there is a local maximum value at \left({\pi \over 2} + 2\pi m,{\pi \over 2} + 2\pi n\right); the local maximum value is z = 2. For integers m and n, there is a local minimum value at \left( - {\pi \over 2} + 2\pi m, - {\pi \over 2} + 2\pi n\right); the local minimum value is z = -2. For integers m and n, \left( - {\pi \over 2} + 2\pi m,{\pi \over 2} + 2\pi n, 0\right) and \left({\pi \over 2} + 2\pi m, - {\pi \over 2} + 2\pi n, 0\right) are saddle points of z.
z=f( x,y) =x^{2}-9y+y^{2}
In Problems 19 and 20, find the absolute maximum and absolute minimum of f on the domian D.
z=f( x,y) = x^{2}- 2xy +2y, D: 0\le x \le 3, 0\le y\le 4.
The absolute maximum value is 9; the absolute minimum value is -7.
z=f( x,y) =3xy^{2}, D: x^{2}+y^{2}\leq 9.
In Problems 21–24, use Lagrange multipliers to find the maximum and minimum values of f subject to the constraints g(x,y)=0.
z=f( x,y) =5x^{2}+3y^{2}+xy, g( x,y) =2x-y-20=0
There is a minimum value at \left({{130} \over {19}},-{{ 120} \over {19}}\right); the minimum value is z = {{5900} \over {19}}.
z=f( x,y) =x\sqrt{y}, g( x,y) =2x+y-3000=0
z=f( x,y) =x^{2}+y^{2}, g( x,y) =2x+y-4=0
There is a minimum value at \left({8 \over 5},{4 \over 5}\right); the minimum value is z={{16} \over 5}.
z=f( x,y) =xy^{2}, g( x,y)=x^{2}+y^{2}-1=0
Heat Transfer A metal plate is placed on the xy-plane in such a way that the temperature T at any point (x,y) is given by T = e^{y}(\sin x + \sin y)^{\circ} C.
Maximizing Profit A company produces two products at a total cost C( x,y) =x^{2}+200x+y^{2}+100y-xy, where x and y represent the units of each product sold. The revenue function is R( x,y) =2000x-2x^{2}+100y-y^{2}+xy.
901
Maximizing Profit A manufacturer introduces a new product with a Cobb–Douglas production function of P( x,y) = 10K^{0.3}L^{0.7}, where K represents the units of capital and L the units of labor needed to produce P units of the product. A total of $51,000 has been budgeted for production. Each unit of labor costs the manufacturer $100 and each unit of capital costs $50.
Volume Find the volume of the largest rectangular solid that can be inscribed in the interior of the surface \dfrac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1 if the sides of the solid are parallel to the axes.
Use Lagrange multipliers to find the points on the surface xyz=1 closest to the origin.
The points (1, 1, 1), (-1, 1,-1), (-1,-1, 1), and (1, -1, -1) are all closest to the origin.
Volume Use Lagrange multipliers to maximize the volume of a rectangular solid that has three faces on the coordinate planes and one vertex on the plane \dfrac{x}{a} +\dfrac{y}{b} +\dfrac{z}{c} =1,a>0,b>0,c>0.
Minimizing Cost The base of a rectangular box costs five times as much as do the other five sides. Use Lagrange multipliers to find the proportions of the dimensions for the cheapest possible box of volume V.
A box with a square base with each side measuring \sqrt[3]{\dfrac{V}{3}} and a height of 3\sqrt[3]{\dfrac{V}{3}} will minimize the total cost of the box.
Find the extreme values of f(x,y,z)=xyz subject to the constraints x^{2}+y^{2}=1 and y=3z.