optimization....

Question

Minimize the function  M=4x^2+4y^2  subject to  2x+y=32

and find the values of x and y at which M is minimized.

The minimum value of M is ___

and it occurs when x is ____

and y is ____

Answer 1

I hope this helps you on your quest.

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Answer 2

I found the same answer as Yefim but in a different way. I took y = 32-2x and substituted into the original equation:

M=4x² + 4y² = 4x² + 4(32-2x)² = 20x² - 512x + 4096

Take the derivative of M which, when set equal to zero, will show you relative maxes and mins.

M' = 40x - 512

At M' = 0, x = 64/5. M' is negative to the left of 64/5 and positive to the right, so the M value at x=64/5 will be a relative and absolute min (absolute because M' will equal zero only once).

Plug that x-value into the original equation y = 32 - 2x to get the y-value and plug those into the formula for M to get the minimum.

Answer 3

Lagrang function L = (4x² + 4y²) - k(2x + y - 32);

L_x = 8x - 2k = 0; L_y = 8y - k = 0; L_k = - 2x - y + 32 = 0; x = k/4, y = k/8; - k/2 - k/8 = - 32; 5k/8 = 32; k = 51.2

x = 12.8, y = 6.4

M(12.8, 6.4) = 819.2

Let x = 12 y = 8 also from 2x + y = 32; L(12, 8) = 832 > 819.2

So, at x = 12.8, y = 6.4 we ave minimum M_min = 819.2