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3 years ago

Maximize Q=xy^2, where x and y are positive numbers, such that x + y^2=10

1 answer:

8 0

X+y^2=10
minus x from both sides
y^2=10-x
sub (10-x) for y^2 in other equation
Q=x(10-x)
Q=10x-x^2
now find the maximum value
take the derititive
dQ/dx=10-2x
it is zero at x=5
below that, it is positive
after 5, it is negative
max at x=5

solve for y
y^2=10-x
y^2=10-5
y^2=5
sqrt both sides
y=√5

x=5
y=√5
the max value is 25

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Write the polynomial as a square of a binomial or as an expression opposite to a square of a binomial:

spayn [35]

Answer:

A) $0.25x^2-0.6xy+0.36y^2=\left(0.5x-0.6y\right)^2$

B) $-a^2+0.6a-0.09=-\left(10a-3\right)^2$

C) $\frac{9a^4}{16}+a^3+\frac{4a^2}{9}=a^2(\left(9a+8\right)^2)$

D) $-16m^2-24mn -9n^2=-\left(4m+3n\right)^2$

Step-by-step explanation:

The square of a binomial is the sum of: the square of the first terms, twice the product of the two terms, and the square of the last term.

$(a+b)^2 = a^2 + 2ab + b^2\\\\(a-b)^2 = a^2 - 2ab + b^2$

To find the square of the binomial of the following polynomials you must:

A) $0.25x^2-0.6xy+0.36y^2$

Apply radical rule: $a=\left(\sqrt{a}\right)^2$

$0.25=\left(\sqrt{0.25}\right)^2\\0.36=\left(\sqrt{0.36}\right)^2$

$\left(\sqrt{0.25}\right)^2x^2-0.6xy+\left(\sqrt{0.36}\right)^2y^2$

Apply exponent rule: $a^mb^m=\left(ab\right)^m$

$\left(\sqrt{0.25}\right)^2x^2=\left(\sqrt{0.25}x\right)^2\\\left(\sqrt{0.36}\right)^2y^2=\left(\sqrt{0.36}y\right)^2$

$\left(\sqrt{0.25}x\right)^2-0.6xy+\left(\sqrt{0.36}y\right)^2$

Rewrite $0.6xy$ as $2\cdot \:0.5x\cdot \:0.6y$

$\left(\sqrt{0.25}x\right)^2-2\cdot \:0.5x\cdot \:0.6y+\left(\sqrt{0.36}y\right)^2$

Apply perfect square formula: $\left(a-b\right)^2=a^2-2ab+b^2$

$a=0.5x,\:b=0.6y$

$\left(\sqrt{0.25}x\right)^2-2\cdot \:0.5x\cdot \:0.6y+\left(\sqrt{0.36}y\right)^2=\left(0.5x-0.6y\right)^2$

B) $-a^2+0.6a-0.09$

Multiply both sides by 100

$-a^2\cdot \:100+0.6a\cdot \:100-0.09\cdot \:100\\-100a^2+60a-9$

Factor out common term -1

$-\left(100a^2-60a+9\right)$

Break the expression into groups and factor out common terms

$-(\left(100a^2-30a\right)+\left(-30a+9\right))\\-(10a\left(10a-3\right)-3\left(10a-3\right))\\-(\left(10a-3\right)\left(10a-3\right))\\-\left(10a-3\right)^2$

C) $\frac{9a^4}{16}+a^3+\frac{4a^2}{9}$

Apply exponent rule: $a^{b+c}=a^ba^c$

$a^3=aa^2\\a^4=a^2a^2$

$\frac{9a^2a^2}{16}+aa^2+\frac{4a^2}{9}$

Factor out common term $a^2$

$a^2\left(\frac{9a^2}{16}+a+\frac{4}{9}\right)$

Factor $\frac{9a^2}{16}+a+\frac{4}{9}\right$

Find the Least Common Multiplier (LCM) of 16, 9 which is 144.

Multiply by LCM

$\frac{9a^2}{16}\cdot \:144+a\cdot \:144+\frac{4}{9}\cdot \:144\\81a^2+144a+64$

$81a^2+144a+64=\left(9a\right)^2+2\cdot \:9a\cdot \:8+8^2$

Apply perfect square formula: $\left(a+b\right)^2=a^2+2ab+b^2$

$a=9a,\:b=8$

$81a^2+144a+64=\left(9a+8\right)^2$

$\frac{9a^4}{16}+a^3+\frac{4a^2}{9}=a^2(\left(9a+8\right)^2)$

D) $-16m^2-24mn -9n^2$

Factor out common term -1

$-\left(16m^2+24mn+9n^2\right)$

Break the expression into groups and factor out common terms

$\left(16m^2+12mn\right)+\left(12mn+9n^2\right)\\4m\left(4m+3n\right)+3n\left(4m+3n\right)\\\left(4m+3n\right)\left(4m+3n\right)\\-\left(4m+3n\right)\left(4m+3n\right)\\-\left(4m+3n\right)^2$

$-16m^2-24mn -9n^2=-\left(4m+3n\right)^2$

3 0

3 years ago

Luke has 1/4

zvonat [6]

2/3
because 14/3= 2.66666666
14-2.666666= 9.33333333
9.333333/14= 2/3

6 0

3 years ago

A researcher is comparing the effectiveness of three devices designed to help people who snore. There are 210 people who snore p

hjlf

Answer:

the answer is 000-210

Step-by-step explanation:

edge 2021

8 0

3 years ago

Ms. Wendie puts 4 chocolate candies in a bag to give each of her students on the first day of class. She puts a total of 76 choc

Alchen [17]

This scenario is all about creating equal groups of 4 candies. Because we know the total number of candies that were used, the 4 is a part and we are missing the other part(groups). We can think of it as how many groups of 4 will result in 76. This is shown in Choice A.

8 0

4 years ago

The time to complete a standardized exam is approximately normal with a mean of 70 minutes and a standard deviation of 10 minus.

AnnyKZ [126]

Answer:

Step-by-step explanation:

Given that the time to complete a standardized exam is approximately normal with a mean of 70 minutes and a standard deviation of 10 minutes.

P(completing exam before 1 hour)

= P(less than an hour) = P(X<60)

=P(Z< $\frac{60-70}{10} =-1$ )

=0.5-0.34=0.16

i.e. 16% of students completed the standardized exam.

7 0

3 years ago