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2 years ago

The cost of a ball is $0.58. David pays for the ball with

Mathematics

2 answers:

luda_lava [24]2 years ago

8 0

Answer:

Step-by-step explanation:

It's A just trust me I'm not doing the work lol

vovangra [49]2 years ago

4 0

Answer:

The answer would be A. $.042

Step-by-step explanation:

For you to find out the answer to how much change he should get back, subtracting the two numbers will get you the answer

1.00 - .58 = .42

Feel free to let me know if you need more help. :)

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2 years ago

Find the extreme values of the function f(x, y) = 4x2 + 6y2 on the circle x2 + y2 = 1.

julia-pushkina [17]

Answer with Step-by-step explanation:

We are given that

$f(x,y)=4x^2+6y^2$

Let g(x,y)= $x^2+y^2=1$

We have to find the extreme values of the given function

$\nabla f(x,y)$

$\nabla g(x,y)=$

Using Lagrange multipliers

$\nabla f(x,y)=\lambda \nabla g(x,y)$

$f_x=\lambda g_x$

$8x=\lambda 2x$

Possible value x=0 or $\lambda=4$

If x=0 then substitute the value in g(x,y)

Then, we get $y=\pm 1$

$f_y=\lambda g_y$

$12y=\lambda 2y$

If $\lambda=4$ and substitute in the equation

Then , we get possible value of y=0

When y=0 substitute in g(x,y) then we get

$x=\pm 1$

Hence, function has possible extreme values at points (0,1),(0,-1), (1,0) and (-1,0).

$f(0,1)=6$

$f(0,-1)=6$

$f(1,0)=4$

$f(-1,0)=4$

Therefore, the maximum value of f on the circle $x^2+y^2=1$ is $f(0,\pm1)=6$ and minimum value of $f(\pm1,0)=4$

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2 years ago