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oksano4ka [1.4K]

2 years ago

8

Javier is comparing two checking accounts. One has a monthly fee of $10 and a per-check fee of $0.10, and the other has a monthl

y fee of $12 and a per-check fee of $0.05. What is the minimum number of checks Javier needs to write for the second bank to be a better option?

2 answers:

almond37 [142]2 years ago

8 0

When Javier writes 40 on both accounts they cost him the same amount of money so it should be 41.

Javier needs to write 41 checks for the second bank to beat better

mel-nik [20]2 years ago

7 0

Answer:

41 is the answer for this question. Hope this helps.

Step-by-step explanation:

(apex)

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Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

aliya0001 [1]

The Lagrangian

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^4+y^4+z^4-13)$

has critical points where the first derivatives vanish:

$L_x=2x+4\lambda x^3=2x(1+2\lambda x^2)=0\implies x=0\text{ or }x^2=-\dfrac1{2\lambda}$

$L_y=2y+4\lambda y^3=2y(1+2\lambda y^2)=0\implies y=0\text{ or }y^2=-\dfrac1{2\lambda}$

$L_z=2z+4\lambda z^3=2z(1+2\lambda z^2)=0\implies z=0\text{ or }z^2=-\dfrac1{2\lambda}$

$L_\lambda=x^4+y^4+z^4-13=0$

We can't have $x=y=z=0$ , since that contradicts the last condition.

(0 critical points)

If two of them are zero, then the remaining variable has two possible values of $\pm\sqrt[4]{13}$ . For example, if $y=z=0$ , then $x^4=13\implies x=\pm\sqrt[4]{13}$ .

(6 critical points; 2 for each non-zero variable)

If only one of them is zero, then the squares of the remaining variables are equal and we would find $\lambda=-\frac1{\sqrt{26}}$ (taking the negative root because $x^2,y^2,z^2$ must be non-negative), and we can immediately find the critical points from there. For example, if $z=0$ , then $x^4+y^4=13$ . If both $x,y$ are non-zero, then $x^2=y^2=-\frac1{2\lambda}$ , and

$xL_x+yL_y=2(x^2+y^2)+52\lambda=-\dfrac2\lambda+52\lambda=0\implies\lambda=\pm\dfrac1{\sqrt{26}}$

$\implies x^2=\sqrt{\dfrac{13}2}\implies x=\pm\sqrt[4]{\dfrac{13}2}$

and for either choice of $x$ , we can independently choose from $y=\pm\sqrt[4]{\frac{13}2}$ .

(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)

If none of the variables are zero, then $x^2=y^2=z^2=-\frac1{2\lambda}$ . We have

$xL_x+yL_y+zL_z=2(x^2+y^2+z^2)+52\lambda=-\dfrac3\lambda+52\lambda=0\implies\lambda=\pm\dfrac{\sqrt{39}}{26}$

$\implies x^2=\sqrt{\dfrac{13}3}\implies x=\pm\sqrt[4]{\dfrac{13}3}$

and similary $y,z$ have the same solutions whose signs can be picked independently of one another.

(8 critical points)

Now evaluate $f$ at each critical point; you should end up with a maximum value of $\sqrt{39}$ and a minimum value of $\sqrt{13}$ (both occurring at various critical points).

Here's a comprehensive list of all the critical points we found:

$(\sqrt[4]{13},0,0)$

$(-\sqrt[4]{13},0,0)$

$(0,\sqrt[4]{13},0)$

$(0,-\sqrt[4]{13},0)$

$(0,0,\sqrt[4]{13})$

$(0,0,-\sqrt[4]{13})$

$\left(\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2},0\right)$

$\left(\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)$

$\left(-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2},0\right)$

$\left(-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)$

$\left(\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)$

$\left(\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)$

$\left(-\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)$

$\left(-\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)$

$\left(0,\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)$

$\left(0,\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)$

$\left(0,-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)$

$\left(0,-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)$

$\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)$

$\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)$

$\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)$

$\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)$

$\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)$

$\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)$

$\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)$

$\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)$

5 0

3 years ago

2/3p+3. p=3/5<br> What is the Answer to what

Vedmedyk [2.9K]

Step-by-step explanation:

2/3p+3

where p=3/5

substitute p into 2/3p + 3

2/3(3/5) + 3

2/(9/5) + 3

10/9 + 3

10+3)/9

13/9

1 4/9

3 0

3 years ago

What is the length of the hypotenuse of a right triangle whose legs have lengths of 5 and 12?

ruslelena [56]

It would be 13... if I did my math right.

6 0

2 years ago

Read 2 more answers

The area of the vegetable garden is 0.4 of the community area of 4.5 x 6.2 . What is the area of the vegetable garden?

Gala2k [10]

Answer:

11.16 units²

Step-by-step explanation:

Area of community area=4.5×6.2= 27.9 units²

Area of vegetable garden=27.9×0.4=11.16 units²

5 0

2 years ago

Work out 3/5 of 900

Mars2501 [29]

Answer:

540

Step-by-step explanation:

1. We assume, that the number 900 is 100% - because it's the output value of the task.

2. We assume, that x is the value we are looking for.

3. If 900 is 100%, so we can write it down as 900=100%.

4. We know, that x is 3.5% of the output value, so we can write it down as x=3.5%.

5. Now we have two simple equations:

1) 900=100%

2) x=3.5%

where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that:

900/x=100%/3.5%

6. Now we just have to solve the simple equation, and we will get the solution we are looking for.

7. Solution for what is 3.5% of 900

900/x=100/3.5

(900/x)*x=(100/3.5)*x - we multiply both sides of the equation by x

900=28.571428571429*x - we divide both sides of the equation by (28.571428571429) to get x

900/28.571428571429=x

31.5=x

x=31.5

now we have:

3.5% of 900=31.5

3 0

3 years ago