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2 years ago

Find A and B given that the function y=Ax−−√+Bx−−√ has a minimum value of 8 at x=1.

Mathematics

1 answer:

docker41 [41]2 years ago

6 0

Answer:

$A=\pm 2\\B=\pm 2$

Step-by-step explanation:

Given that

$[tex]8=\sqrt{A}+\sqrt{A}\\\\8=2\sqrt{A}\\\\\sqrt{A}=4\\\\A=\pm 2\\\\which\,\, implies\\ \\B=\pm 2$ ---(1)\\[/tex]

To find minimum differentiate both sides w.r.to x

$\frac{dy}{dx}=\frac{d}{dx}\sqrt{Ax}+\frac{d}{dx}\sqrt{Bx}\\\\\frac{dy}{dx}=\frac{A}{\sqrt{Ax}}+\frac{B}{\sqrt{Bx}}\\\\At\,\, minima\,\,\frac{dy}{dx}=0\\\\\frac{A}{\sqrt{Ax}}+\frac{B}{\sqrt{Bx}}=0\\\\\frac{A\sqrt{Bx}+B\sqrt{Ax}}{x\sqrt{AB}}=0\\\\\frac{\sqrt{x}(A\sqrt{B}+B\sqrt{A})}{x\sqrt{AB}}=0\\\\\frac{(A\sqrt{B}+B\sqrt{A})}{\sqrt{ABx}}=0\\\\(A\sqrt{B}+B\sqrt{A})=0\\\\A\sqrt{B}=-B\sqrt{A}\\\\\sqrt{A}=-\sqrt{B}\\\\A=B--(2)\\\\$

Substituting (2) in (1) at x=1, y=8

$8=\sqrt{A}+\sqrt{A}\\\\2\sqrt{A}=8\\\\\sqrt{A}=4\\\\A=\pm 2\\\\which\,\, implies\\\\B=\pm 2$

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Demarco and Tanya have received information about three separate mortgage offers. In two or three paragraphs, describe your reco

Mars2501 [29]

Answer:

Mortgage option (3) would be best suited for them.

Step-by-step explanation:

Mortgage option (1) and (2) are more or less the same since, since even if Damarco and Tanya down payments $34,000 (20% of the purchase price), they need to pay the interest for 30 years for both of the cases and even if he pays about $750 monthly (as for option (1)) or about $ 9000 annually (as for option (2)) both may actually be more or less the same amount since, the annual rate of interest in (2) may increase from the initial rate of 3.5% (but it is very unlikely to increase to over 5%) and option (1) has an annual fixed rate of interest of 4.25%.

Now, in the option (3) the interest is to be paid for 8 years and the annual rate of interest is also relatively low (only 4%) and if they pay about $18,000 annually with a down-payment of $ 34,000 and repay the rest of the amount at the end of 8 years,(which would be less than $ 35,000) they can easily clear their mortgage. Hence, for option (3) they would need to pay lowest total amount and for lowest time to clear the mortgage among the three options. Hence, this would be best suited option for them.

6 0

2 years ago

Problem 5 (4+4+4=12) We roll two fair 6-sided dice. Each one of the 36 possible outcomes is assumed to be equally likely. 1) Fin

tekilochka [14]

Answer:

$p(b) = \frac{1}{6}$

$p(k) = \frac{1}{3}$

$P(a) = \frac{1}{3}$

Step-by-step explanation:

Generally when two fair 6-sided dice is rolled the doubles are

(1 1) , ( 2 2) , (3 3) , (4 4) , ( 5 5 ), (6 6)

The total outcome of doubles is N = 6

The total outcome of the rolling the two fair 6-sided dice is

n = 36

Generally the probability that doubles (i.e., having an equal number on the two dice) were rolled is mathematically evaluated as

$p(b) = \frac{N}{n}$

$p(b) = \frac{6}{36}$

$p(b) = \frac{1}{6}$

Generally when two fair 6-sided dice is rolled the outcome whose sum is 4 or less is

(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)

Looking at this outcome we see that there are two doubles present

The conditional probability that doubles were rolled is mathematically represented as

$p(k) = \frac{2}{6}$

$p(k) = \frac{1}{3}$

Generally when two fair 6-sided dice is rolled the number of outcomes that would land on different numbers is L = 30

And the number of outcomes that at least one die is a 1 is W = 10

The conditional probability that at least one die is a 1 is mathematically represented as

$P(a) = \frac{W}{L}$

=> $P(a) = \frac{10}{30}$

=> $P(a) = \frac{1}{3}$

3 0

2 years ago

Two samples each of size 20 are taken from independent populations assumed to be normally distributed with equal variances. The

Harlamova29_29 [7]

Answer:

$t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923$

The degrees of freedom are

$df=20+20-2=38$

And the p value is given by:

$p_v =2*P(t_{38}>2.923) =0.0058$

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

We have the following data given:

$n_1 =20$ represent the sample size for group 1

$n_2 =20$ represent the sample size for group 2

$\bar X_1 =43.5$ represent the sample mean for the group 1

$\bar X_2 =40.1$ represent the sample mean for the group 2

$s_1=4.1$ represent the sample standard deviation for group 1

$s_2=3.2$ represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

$\S^2_p =\frac{(20-1)(4.1)^2 +(20 -1)(3.2)^2}{20 +20 -2}=13.525$

And the deviation would be just the square root of the variance:

$S_p=3.678$

The statistic is givne by:

$t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923$

The degrees of freedom are

$df=20+20-2=38$

And the p value is given by:

$p_v =2*P(t_{38}>2.923) =0.0058$

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2 years ago

A person is watching a boat from the top of a lighthouse. The boat is approaching the lighthouse directly. When first noticed, t

xxMikexx [17]

If we let
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y as the distance between the boat and the lighthouse.

Then, we have:
tan 18°33' = 200 / (x + y)
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tan 51°33' = 200 / y

Solving for y in the second equation:
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Rearranging the first equation and substituting y
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x = 458.81 ft

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Korvikt [17]

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2 years ago

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