1. If f(x, y, z) = ye^x + xln(z), then fxyz(x, y, z) is equal to a. (2yz+x)/z b.

Question

1. If f(x, y, z) = ye^x + xln(z), then fxyz(x, y, z) is equal to a. (2yz+x)/z b. -1/z^2 c. -x/z^2 d. (x-zy)/z
2. Find the positive numbers x, y, z such that x+y+z=20 and xyz^2 is a maximum.
3. The region in xy plane is bounded above y=4-x^2 and bounded below by the x axis. Find the volume under the graph of f(x, y) = x + 2y + 3 and above the region. 1. If f(x, y, z) = ye^x + xln(z), then fxyz(x, y, z) is equal to a. (2yz+x)/z b. -1/z^2 c. -x/z^2 d. (x-zy)/z
2. Find the positive numbers x, y, z such that x+y+z=20 and xyz^2 is a maximum.
3. The region in xy plane is bounded above y=4-x^2 and bounded below by the x axis. Find the volume under the graph of f(x, y) = x + 2y + 3 and above the region. a. (2yz+x)/z b. -1/z^2 c. -x/z^2 d. (x-zy)/z
2. Find the positive numbers x, y, z such that x+y+z=20 and xyz^2 is a maximum.
3. The region in xy plane is bounded above y=4-x^2 and bounded below by the x axis. Find the volume under the graph of f(x, y) = x + 2y + 3 and above the region.

Explanation / Answer

f ( x,y,z) = yex + x ln z
f xyz (x,y,z) =
f x (x,y,z) = yex + ln z
fxy ( x,y,z) = ex
fxyz ( x,y,z) = 0;

Thus, fxyz = 0;

2) x + y + z = 20; and x y z2 has to be maximised;
xyz2 = x * y * z * z ;
Thus, the product is made up of 4 terms, for the product to be maximum, each of these terms should be the same;
Thus, for maximum value of the product: x = y = z = z;
There are 4 terms, so 20 should be divided into 4 parts ; 20/4 = 5;
Thus, x = 5; y = 5; z = 5*2 ( in proportion with their powers)
Thus, for maximum value of xyz2 , x=5; y= 5 and z=10;  
xyz2 = 5 * 5 * 102 = 25*100 = 2500;

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