calculus/algebra problem?

actually, it may be an algebra problem, but i need to know how to do this in order to solve for local maxima and minima.

fx=4x+3y+2 and fy=3x+8y-10

I need to find x and y by substitution, I suppose, but how? If the coefficients of x were more agreeable I could solve by elimination but I don't think I can in this problem. Thanks!

Comments

  • If you just want to know the x and y for those 2 equations (I hope they are not from the same question XD).

    f(x)=4x+3y+2

    Since we have 2 unknowns we have to find one of them first. Let's try and find x (cuz I feel like it XD, but you can try and solve for y and see if you get the same answer I did at the very end for fun. (which you should)), using substitution method.

    f(x)=4x+3y+2

    4x=3y+2 (No sign change to 4x... an exception for if you want to find an answer of 2 unknowns)

    I brought the 4x to the other side and remember, when you bring numbers to the OTHER SIDE... it's sign DOESNT change. So its still 4x now. You also have to divide the OTHER SIDE... by 4, cuz what you do on one side, you must do the same on the other.

    4x/4 = 3y/4 + 2/4

    4x/4 = 3y + 2/4 (Since 3y/4 + 2/4 have a common denominator of 4, you can put 3y and 2 together if you want. But I won't because it's easier to show you how to do this question without putting them together).

    So now you have:

    x = 3y/4 + 1/2 <-- was 2/4, but simplified.

    But x is positive... so no need to reverse all signs:

    x= 3y/4 + 1/2

    See, now we know what x is equal to! Hooray! But obviously, it has a y in it cuz in the original equation, there were 2 unknowns. Let's call the step that we just did to find x, the 'equation of constraint'.

    Like any other question with 1 unknown, you will try to find what the unknown is equal to. If you imagine that this 2 unknown equation has 1 unknown ( in this case we chose x o be our unknown), you can just solve for that unknown by imagining that you are finding the answer to an equation with only 1 unknown (so basically, partially pretend that the y doesnt exist like I just did).

    Now plug that x value into the original equation, like you would do with any other algebraic question.

    (x = 3y/4 + 1/2)

    f(x)=4x+3y+2

    f(x)=4(3y/4 + 1/2) + 3y + 2 (Sub in x and Mulitply)

    f(x)=(12y/4 + 4/2) + 3y + 2 (Common Denominator time)

    f(x)=12y/4 + 3y + 4/2 + 2 (if you want, put the like terms together, and simplify 12y/4)

    f(x)=6y + 4

    6y = 4 (Put it on the other side)

    y= 4/6

    Now that you found an answer WITHOUT a variable in it, sub it in the original equation again to find the OTHER variable... which is x again because we had a temporary answer for it due to the variable it had.

    f(x)=4x+3y + 2

    f(x)=4x+3(4/6) + 2

    -4x=12/18 + 2

    -4x=12/18 + 36/18

    4x = -12/18 - 36/18

    x = (-48/18)/4

    x = -2/3

    ***************

    I just did the substitution method for you, I dont know which or what minima and maxima your finding for... the function you gave in your question? Anyways...

    ***************

    Try to use this sub method for fy=3x+8y-10 on your own now.

    The answer is below (to see if it matched up with mine)

    f(y)=3x+8y-10

    8y = 3x - 10

    y= 3x/8 - 10/8

    f(y)=3x + 8y-10

    f(y)=3x + 8(3x/8 - 10/8) - 10

    f(y)=3x + 24x/8 - 80/8 - 10

    f(y)=3x + 3x - 10 - 10

    -6x = -20

    x = 20/6

    x = 10/3

    f(y)=3x + 8y-10

    f(y)=3(10/3) + 8y -10

    f(y)=30/3 + 8y- 10

    f(y)=10 + 8y - 10

    f(y)= 8y

    -8y = 0

    8y = 0

    y = 0

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