# How to find critical points and the highest point on the graph of #4x^2 + 4xy+7y^2=15#?

Find

so we have (in the original equation):

Here is the graph: graph{4x^2 + 4xy+7y^2=15 [-4.385, 4.384, -2.19, 2.195]}

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To find critical points on the graph of (4x^2 + 4xy + 7y^2 = 15), we first need to take partial derivatives with respect to (x) and (y) and then solve the resulting system of equations.

Taking the partial derivative with respect to (x) gives us: (8x + 4y = 0), and with respect to (y) gives us: (4x + 14y = 0).

Solving the system of equations simultaneously, we find that (x = -\frac{7}{3}) and (y = \frac{7}{6}).

Substituting these values back into the original equation, we find that the critical point is (\left(-\frac{7}{3}, \frac{7}{6}\right)).

To find whether this critical point corresponds to a maximum or minimum, we can use the second derivative test.

The second partial derivatives with respect to (x) and (y) are (8) and (14) respectively.

The determinant of the Hessian matrix ((H)) is (D = (8)(14) - (4)^2 = 112 - 16 = 96).

Since (D > 0) and the second partial derivative with respect to (x) is positive, the critical point corresponds to a minimum.

Therefore, the highest point on the graph corresponds to the critical point (\left(-\frac{7}{3}, \frac{7}{6}\right)).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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