Determine the global extreme values of the function f(x, y)=x3+x2y+7y2
f(x, y)=x3+x2y+7y2 on
x, y≥0, x+y≤3.
x, y≥0, x+y≤3. (Use symbolic notation and fractions where needed.)

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3

can you rephrase the question ? it is unclear : -)

step-by-step explanation:

keene f. ffnfnfnfnfnfnfn f bfnfnfnfnfj

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