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Ksenya-84 [330]

2 years ago

6

Hello, could you help me understand how to solve this question. Thank you. Also could you explain what does "How does this ratio

compare to the scale factor?"

1 answer:

VikaD [51]2 years ago

5 0

Answer:

Step-by-step explanation:

I think the quotient of the ratio is the scale factor.

C'A' / CA = 8/4 = 2

So since the scale factor is 2 it means that the triangle on the right has dimensions that are twice those of the triangle on the left.

If you need more, let me know.

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You're looking for the extreme values of $x^2+3y^2+13x$ subject to the constraint $x^2+y^2\le1$ .

The target function has partial derivatives (set equal to 0)

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$\dfrac{\partial(x^2+3y^2+13x)}{\partial y}=6y=0\implies y=0$

so there is only one critical point at $\left(-\dfrac{13}2,0\right)$ . But this point does not fall in the region $x^2+y^2\le1$ . There are no extreme values in the region of interest, so we check the boundary.

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$T'(u)=-13\sin u+4\sin u\cos u=\sin u(4\cos u-13)=0$

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but $|\cos u|\le1$ for all $u$ , so this case yields nothing important.

At these critical points, we have temperatures of

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so the plate is hottest at (1, 0) with a temperature of 14 (degrees?) and coldest at (-1, 0) with a temp of -12.

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