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  • MarkFL's Avatar
    Today, 19:21
    MarkFL replied to a thread Lagrange Multipliers 2 in Calculus
    I agree that the point $(2,2)$, is the only one that meets all criteria. Now we need to compare the value of $f$ at another point on the constraint,...
    6 replies | 47 view(s)
  • MarkFL's Avatar
    Today, 19:08
    MarkFL replied to a thread Lagrange Multipliers in Calculus
    I agree that of the 3 critical points, $(1,1)$ is the only one in quadrant I. Now, we know this is either a maximum or a minimum, and to determine...
    6 replies | 40 view(s)
  • MarkFL's Avatar
    Today, 11:15
    MarkFL replied to a thread Factoring...6 in Pre-Calculus
    It might be more clear to state something like the following: The difference of cubes formula states: p^3-q^3=(p-q)\left(p^2+pq+q^2\right) ...
    4 replies | 56 view(s)
  • MarkFL's Avatar
    Today, 10:47
    MarkFL replied to a thread Lagrange Multipliers 2 in Calculus
    Consider: e^u=0 What do you get when solving for $u$? Okay, you correctly found $x^2=y^2$...what do you get when you substitute for...
    6 replies | 47 view(s)
  • MarkFL's Avatar
    Today, 10:39
    MarkFL replied to a thread Lagrange Multipliers in Calculus
    What I would do is use the constraint to determine $y=2-x$. Now substitute for $y$ in both equations you mentioned, and solve for $x$, then your...
    6 replies | 40 view(s)
  • MarkFL's Avatar
    Today, 02:30
    This is a calculus question...please don't continue to post calculus questions in other forums. If given: ...
    3 replies | 78 view(s)
  • MarkFL's Avatar
    Yesterday, 23:12
    MarkFL replied to a thread Lagrange Multipliers 2 in Calculus
    If you solve both equations for $\lambda$ and then equate the results, you obtain: \frac{ye^{xy}}{2x}=\frac{xe^{xy}}{2y} Multiply through by 2:...
    6 replies | 47 view(s)
  • MarkFL's Avatar
    Yesterday, 21:24
    MarkFL replied to a thread Lagrange Multipliers in Calculus
    Okay, so what this implies is: \frac{x}{\sqrt{6-x^2-y^2}}=\frac{y}{\sqrt{6-x^2-y^2}} Cross-multiply: x\sqrt{6-x^2-y^2}=y\sqrt{6-x^2-y^2} ...
    6 replies | 40 view(s)
  • MarkFL's Avatar
    Yesterday, 16:18
    MarkFL replied to a thread The Distance Across in Geometry
    Using the Pythagorean theorem, we find: \overline{MK}=\sqrt{(\sqrt{2}a)^2+(\sqrt{2}b)^2}=\sqrt{2\left(a^2+b^2\right)} Now, we know that...
    6 replies | 63 view(s)
  • MarkFL's Avatar
    Yesterday, 06:10
    MarkFL replied to a thread [SOLVED] Minimum of function under constraint in Calculus
    I would use W|A: W|A - optimize 2x+y subject to xy=18
    8 replies | 249 view(s)
  • MarkFL's Avatar
    Yesterday, 04:29
    The objective function is linear, so it describes a plane, and so I don't believe there will be any saddle points, or in fact any critical points...
    4 replies | 60 view(s)
  • MarkFL's Avatar
    Yesterday, 04:04
    MarkFL replied to a thread Boat direction in Calculus
    Is this a calculus-based physics course?
    6 replies | 91 view(s)
  • MarkFL's Avatar
    Yesterday, 01:37
    What I would do is observe that we have cyclical symmetry, that is we may interchange $x_1$ and $x_2$ with no change in either the objective function...
    4 replies | 60 view(s)
  • MarkFL's Avatar
    Yesterday, 01:25
    MarkFL replied to a thread The Distance Across in Geometry
    Not quite...it would be \sqrt{2}a and \sqrt{2}b...so what would the diagonal of the rectangle be?
    6 replies | 63 view(s)
  • MarkFL's Avatar
    March 23rd, 2017, 21:17
    MarkFL replied to a thread The Distance Across in Geometry
    I would let $a$ be the length (in cm) of the segments with 1 tick mark, and $b$ be the length (in cm) of the segments with 2 tick marks. And so,...
    6 replies | 63 view(s)
  • MarkFL's Avatar
    March 23rd, 2017, 19:30
    MarkFL replied to a thread Boat direction in Calculus
    Being that it is an optimization problem, on a function in two variables, the way I worked it using the calculus is the most straightforward way I...
    6 replies | 91 view(s)
  • MarkFL's Avatar
    March 23rd, 2017, 12:53
    MarkFL replied to a thread Boat direction in Calculus
    I think what I would do is orient a coordinate system such that south is up and west is to the right. At time $t=0$, have boat A at the origin, and...
    6 replies | 91 view(s)
  • MarkFL's Avatar
    March 23rd, 2017, 11:28
    There are online calculators that will spit out the answer to this question, however if you want genuine help, we need to know what you've tried, or...
    3 replies | 102 view(s)
  • MarkFL's Avatar
    March 23rd, 2017, 03:19
    MarkFL replied to a thread Factoring...3 in Pre-Calculus
    Finding what to multiply together to get an expression. It is like "splitting" an expression into a multiplication of simpler expressions.
    6 replies | 74 view(s)
  • MarkFL's Avatar
    March 23rd, 2017, 02:50
    MarkFL replied to a thread Factoring...3 in Pre-Calculus
    I would say division is the inverse of multiplication, but factoring is certainly related to division. For example, we know: ...
    6 replies | 74 view(s)
  • MarkFL's Avatar
    March 23rd, 2017, 02:42
    MarkFL replied to a thread Factoring...4 in Pre-Calculus
    Looks good. (Yes)
    2 replies | 34 view(s)
  • MarkFL's Avatar
    March 23rd, 2017, 02:32
    MarkFL replied to a thread Factoring...3 in Pre-Calculus
    Perfect. (Yes)
    6 replies | 74 view(s)
  • MarkFL's Avatar
    March 23rd, 2017, 02:24
    MarkFL replied to a thread Factoring...2 in Pre-Calculus
    I was just rewriting that term as a perfect square: 9(ab+c)^2=3^2(ab+c)^2=(3(ab+c))^2
    4 replies | 47 view(s)
  • MarkFL's Avatar
    March 23rd, 2017, 02:16
    MarkFL replied to a thread Factoring...2 in Pre-Calculus
    I would first write the expression as: (2ab)^2-(3(ab+c))^2 Now, factor as the difference of squares, and simplify. :D
    4 replies | 47 view(s)
  • MarkFL's Avatar
    March 23rd, 2017, 00:48
    As I said before, please post calculus questions in our "Calculus" forum, and please don't begin a new thread for the same question. I have moved and...
    7 replies | 81 view(s)
  • MarkFL's Avatar
    March 22nd, 2017, 20:04
    MarkFL replied to a thread Find Value of k in Pre-Calculus
    True, the roots are no given, but you can compute them...we should begin by equating the given quadratic to zero: 2x^2+5xk=0 Using the...
    8 replies | 95 view(s)
  • MarkFL's Avatar
    March 22nd, 2017, 19:49
    MarkFL replied to a thread Find the Value of n in Pre-Calculus
    You want to set the discriminant greater than or equal to zero, since you don't want it to be negative. If the discriminant is equal to zero, then...
    6 replies | 60 view(s)
  • MarkFL's Avatar
    March 22nd, 2017, 19:21
    MarkFL replied to a thread Find Value of k in Pre-Calculus
    No, the roots themselves. Recall, if: ax^2+bx+c=0 then: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} What are the roots of the given quadratic?
    8 replies | 95 view(s)
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