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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

A rectangle with length L and width W is cut into four smaller rectangles by two lines parallel to the sides. Find the maximum and minimum values of the sum of the squares of the areas of the smaller rectangles.

To determine

To find: The maximum and minimum values of the sum of the squares of the areas of the smaller rectangles.

Explanation

Given:

The rectangle with the length L and width W is cut into four smaller rectangles by two lines which are parallel to the sides of the rectangle as shown below in figure 1.

Calculation:

Let, A1,A2,A3,A4 are the areas of four the smaller rectangles.

From figure 1, the areas are determined as,

A1=xyA2=(Lx)yA3=(Lx)(Wy)A4=x(Wy)

The Summation of squares of the smaller rectangles is the function f(x,y) and it is given below,

f(x,y)=A12+A22+A32+A42

Substitute xy for A1 , (Lx)y for A2 , (Lx)(Wy) for A3 and x(Wy) for A4 ,

f(x,y)=(xy)2+[(Lx)y]2+[(Lx)(Wy)]2+[x(Wy)]2=x2y2+(Lx)2y2+(Lx)2(Wy)2+x2(Wy)2=[x2+(Lx)2][y2+(Wy)2]

f(x,y)=[x2+(Lx)2][y2+(Wy)2] (1)

The conditions to find the maximum and minimum values of f(x,y) are

  1. 1. If fxxfyyfxy2>0 at (x,y) then, (x,y) is either maximum or minimum point.
  2. 2. If fxx,fyy<0 at (x,y) then, (x,y) is a maximum point.
  3. 3. If fxx,fyy>0 at (x,y) then, (x,y) is a minimum point.

For stationary point fx(x,y)=fy(x,y)=0

To find stationary point,

First order derivative are calculated as follows,

Differentiate the equation (1) with respect to x .

fx(x,y)=ddx[f(x,y)]=ddx[x2+(Lx)2][y2+(Wy)2]=[2x2(Lx)][y2+(Wy)2]

For stationary point fx(x,y)=0 .

fx(x,y)=[2x2(Lx)][y2+(Wy)2]=0=2x2L+2x=0=4x2L=0=4x=2L

x=2L4x=L2

Differentiate the equation (1) with respect to y .

fy(x,y)=ddy[f(x,y)]=ddy[x2+(Lx)2][y2+(Wy)2]=[x2+(Lx)2][2y2(Wy)]

For stationary point fy(x,y)=0 .

fy(x,y)=[x2+(Lx)2][2y2(Wy)]=0=2y2W+2y=0=4y2W=0=4y=2W

y=2W4y=W2

Second order derivative are calculated as follows,

To obtain fxx , differentiate the fx with respect to x .

fxx=ddx(fx)=ddx{[2x2(Lx)][y2+(Wy)2]}=ddx{[2x2L+2x][y2+(Wy)2]}=ddx{[4x2L][y2+(Wy)2]}

=(40)[y2+(Wy)2]fxx=4[y2+(Wy)2] (2)

To obtain fyy , differentiate the fy with respect to y

fyy=ddy(fy)=ddy{[x2+(Lx)2][2y2(Wy)]}=ddy{[x2+(Lx)2][2y2W+2y]}=ddy{[x2+(Lx)2][4y2W]}

=(40)[x2+(Lx)2]fyy=4[x2+(Lx)2] (3)

To obtain fxy , differentiate fy with respect to x .

fxy=fyx=ddx(fy)=ddy(fx)

fxy=ddx(fy)=ddx{[x2+(Lx)2][2y2(Wy)]}={[2x2(Lx)][2y2(Wy)]}=(2x2L+2x)(2y2W+2y)

fxy=(4x2L)(4y2W) (4)

The condition for either maximum or minimum point is fxxfyyfxy2>0 .

Substitute 4[y2+(Wy)2] for fxx , 4[x2+(Lx)2] for fyy and (4x2L)(4y2W) for fxy ,

=4[y2+(Wy)2]*4[x2+(Lx)2][(4x2L)(4y2W)]2=16[y2+(Wy)2][x2+(Lx)2][(4x2L)2(4y2W)2]=16[y2+W2+y22Wy][x2+L2+x22Lx][(16x2+4L216xL)(16y2+4W216yW)]

Substitute L2 for x and W2 for y ,

=[(L2)2+L2+(L2)22LL2][(W2)2+W2+(W2)22WW2][(16(L2)2+4L216L2L)(16(W2)2+4W216W2W)]

=(L24+L2+L24L2)(W24+W2+W24W2)[(16L24+4L216L22)(16W24+4W216W22)]=(2L24)(2W24)[(4L28L22)(4W28W22)]=(2L24)(2W24)(4L28L22)(4W28W22)=(2L24)(2W24)(4L28L22)(4W28W22)>0

So the point (L2,W2) will be either maximum or minimum point

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