And Find the dimensions of the rectangular box with the volume in the first octa
ID: 2888158 • Letter: A
Question
And Find the dimensions of the rectangular box with the volume in the first octant with one vertex at the origin and the opposite vertex on the ellipsoid 64x^2+4y^2+16z^2=64
X 12.8.47 Find the absolute maximum and minimum values of the following function on the given region R. f(x,y)-27-4x + 8y2 + 7; Determine the absolute maximum value of f on R. Select the correct choice below and, if necessary, il in the answer bo A. The absolute maximum value of f on R is I (Simplify your answer) 0 B. There is no absolute maximum value
Explanation / Answer
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12.8.47)
given f(x,y)=2x2-4x+8y2+7
fx(x,y)=4x-4,fy(x,y)=16y
for critical points ,fx(x,y)=0,fy(x,y)=0
=>4x-4=0, 16y=0
=>x=1, y=0
so (x,y)=(1,0) is the critical point
(1,0) lies in given region (x-1)2+y21
f(1,0)=2*12-4*1+8*02+7
=>f(1,0)=2-4+0+7
=>f(1,0)=5
on the boundary (x-1)2+y2=1
=>y2=1-(x-1)2
=> y=-(1-(x-1)2),y=(1-(x-1)2)
left most point of this circle is (0,0) is and right most point is (2,0)
f(0,0)=2*02-4*0+8*02+7 =7
f(2,0)=2*22-4*2+8*02+7 =7
f(x,y)=2x2-4x+8y2+7 ,y2=1-(x-1)2
=>f(x)=2x2-4x+8(1-(x-1)2)+7
=>f(x)=2x2-4x+8(1-(x2-2x+1))+7
=>f(x)=2x2-4x+8(-x2+2x)+7
=>f(x)=2x2-4x-8x2+16x+7
=>f(x)=-6x2+12x+7
f '(x)=-12x+12
for extrema , f '(x)=0
=>-12x+12=0
=>x=1
y2=1-(x-1)2,x=1
=>y2=1-(1-1)2
=>y2=1
=>y=-1, y=1
two more candidate points are (x,y)=(1,-1),(1,1)
f(1,-1)=2*12-4*1+8*(-1)2+7 =2-4+8+7 =13
f(1,1)=2*12-4*1+8*(1)2+7 =2-4+8+7 =13
so
the absolute maximum value of f on R is 13 which occurs at (1,-1),(1,1)
the absolute minimum value is of f on R is 5 which occurs at (1,0)
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