minimize the function m = 4x² + 4y² 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 $M = 4x^{2}+4y^{2}$, then $M=4x^{2}+4(48 - 2x)^{2}$. Expand $(48 - 2x)^{2}=2304-192x + 4x^{2}$. So $M=4x^{2}+4(2304-192x + 4x^{2})=4x^{2}+9216-768x + 16x^{2}=20x^{2}-768x + 9216$.

Step3: Find the derivative of M with respect to x

$M^\prime(x)=\frac{d}{dx}(20x^{2}-768x + 9216)=40x-768$.

Step4: Set the derivative equal to 0

Set $M^\prime(x)=0$, then $40x-768 = 0$. Solve for x: $40x=768$, so $x=\frac{768}{40}=\frac{96}{5} = 19.2$.

Step5: Find the value of y

Substitute $x = 19.2$ into $y = 48 - 2x$, we get $y=48-2\times19.2=48 - 38.4 = 9.6$.

Step6: Find the minimum value of M

Substitute $x = 19.2$ and $y = 9.6$ into $M = 4x^{2}+4y^{2}$. $M=4\times(19.2)^{2}+4\times(9.6)^{2}=4\times368.64+4\times92.16=4\times(368.64 + 92.16)=4\times460.8=1843.2$.

Answer:

The minimum value of M is $1843.2$, and it occurs when x is $19.2$ and y is $9.6$.