minimize the function m = 2x² + 2y² subject to 2x + y = 48 and find the values of x and y at which m is…

Explanation:

Step1: Express y in terms of x

From $2x + y=48$, we get $y = 48 - 2x$.

Step2: Substitute y into M

Substitute $y = 48 - 2x$ into $M=2x^{2}+2y^{2}$, then $M=2x^{2}+2(48 - 2x)^{2}$. Expand $(48 - 2x)^{2}=48^{2}-192x + 4x^{2}$. So $M=2x^{2}+2(2304-192x + 4x^{2})=2x^{2}+4608-384x + 8x^{2}=10x^{2}-384x + 4608$.

Step3: Find the derivative of M

Differentiate $M = 10x^{2}-384x + 4608$ with respect to x. $M^\prime=20x-384$.

Step4: Set the derivative equal to 0

Set $M^\prime = 0$, so $20x-384 = 0$. Solve for x: $20x=384$, then $x=\frac{384}{20}=19.2$.

Step5: Find the value of y

Substitute $x = 19.2$ into $y = 48 - 2x$. Then $y=48-2\times19.2=48 - 38.4 = 9.6$.

Step6: Find the minimum value of M

Substitute $x = 19.2$ and $y = 9.6$ into $M=2x^{2}+2y^{2}$. $M=2\times(19.2)^{2}+2\times(9.6)^{2}=2\times368.64+2\times92.16=737.28 + 184.32=921.6$.

Answer:

The minimum value of M is 921.6, and it occurs when x is 19.2 and y is 9.6.