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3 years ago

8

Taylor and her friends are going into a bakery. How much would it cost to buy 5 donuts and 1 cookie if each donut costs $2.00 an

d each cookie costs $0.75? How much would it cost to buy 55 donuts and 11 cookies if each donut cost $d and each cookie cost $c?

1 answer:

zhenek [66]3 years ago

4 0

1. $10.75
2. 55d+11c???

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All boxes with a square base, an open top, and a volume of 60 ft cubed have a surface area given by S(x)equalsx squared plus

Karo-lina-s [1.5K]

Answer:

The absolute minimum of the surface area function on the interval $(0,\infty)$ is $S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

The dimensions of the box with minimum surface area are: the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

Step-by-step explanation:

We are given the surface area of a box $S(x)=x^2+\frac{240}{x}$ where x is the length of the sides of the base.

Our goal is to find the absolute minimum of the the surface area function on the interval $(0,\infty)$ and the dimensions of the box with minimum surface area.

1. To find the absolute minimum you must find the derivative of the surface area ( $S'(x)$ ) and find the critical points of the derivative ( $S'(x)=0$ ).

$\frac{d}{dx} S(x)=\frac{d}{dx}(x^2+\frac{240}{x})\\\\\frac{d}{dx} S(x)=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(\frac{240}{x}\right)\\\\S'(x)=2x-\frac{240}{x^2}$

Next,

$2x-\frac{240}{x^2}=0\\2xx^2-\frac{240}{x^2}x^2=0\cdot \:x^2\\2x^3-240=0\\x^3=120$

There is a undefined solution $x=0$ and a real solution $x=2\sqrt[3]{15}$ . These point divide the number line into two intervals $(0,2\sqrt[3]{15})$ and $(2\sqrt[3]{15}, \infty)$

Evaluate S'(x) at each interval to see if it's positive or negative on that interval.

$\begin{array}{cccc}Interval&x-value&S'(x)&Verdict\\(0,2\sqrt[3]{15}) &2&-56&decreasing\\(2\sqrt[3]{15}, \infty)&6&\frac{16}{3}&increasing \end{array}$

An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We can see from the table that f(x) decreases before $x=2\sqrt[3]{15}$ , increases after it, and is defined at $x=2\sqrt[3]{15}$ . So f(x) has a relative minimum point at $x=2\sqrt[3]{15}$ .

To confirm that this is the point of an absolute minimum we need to find the second derivative of the surface area and show that is positive for $x=2\sqrt[3]{15}$ .

$\frac{d}{dx} S'(x)=\frac{d}{dx}(2x-\frac{240}{x^2})\\\\S''(x) =\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(\frac{240}{x^2}\right)\\\\S''(x) =2+\frac{480}{x^3}$

and for $x=2\sqrt[3]{15}$ we get:

$2+\frac{480}{\left(2\sqrt[3]{15}\right)^3}\\\\\frac{480}{\left(2\sqrt[3]{15}\right)^3}=2^2\\\\2+4=6>0$

Therefore S(x) has a minimum at $x=2\sqrt[3]{15}$ which is:

$S(2\sqrt[3]{15})=(2\sqrt[3]{15})^2+\frac{240}{2\sqrt[3]{15}} \\\\2^2\cdot \:15^{\frac{2}{3}}+2^3\cdot \:15^{\frac{2}{3}}\\\\4\cdot \:15^{\frac{2}{3}}+8\cdot \:15^{\frac{2}{3}}\\\\S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2$

2. To find the third dimension of the box with minimum surface area:

We know that the volume is 60 $ft^3$ and the volume of a box with a square base is $V=x^2h$ , we solve for h

$h=\frac{V}{x^2}$

Substituting V = 60 $ft^3$ and $x=2\sqrt[3]{15}$

$h=\frac{60}{(2\sqrt[3]{15})^2}\\\\h=\frac{60}{2^2\cdot \:15^{\frac{2}{3}}}\\\\h=\sqrt[3]{15} \:ft$

The dimension are the base edge $x=2\sqrt[3]{15}\:ft$ and the height $h=\sqrt[3]{15} \:ft$

6 0

3 years ago

The step function f(x) is graphed.

NISA [10]

From the given step function in this item above, we can see that the value of y in the given equation is dependent in the values of x but with certain step-wise arrangement. As we can see, for x = 1, the value of y is equal to 0. Thus, the answer is the first choice.

8 0

3 years ago

Read 2 more answers

Which of the following is the least cost for a television before taxes?

nika2105 [10]

Answer:

G is cheapest at $815.15

Step-by-step explanation:

F-cost is 915

g cost is 85% (100-15=85%) so .85 x 959=815.15

h 1000-100=900

j 925-20=905.

4 0

3 years ago

Find the area of a triangle bounded by the y-axis, the line f(x)=9−2/3x, and the line perpendicular to f(x) that passes through

MaRussiya [10]

Answer:

The area of the triangle is 18.70 sq.units.

Step-by-step explanation:

It is provided that a triangle is bounded by the y-axis, the line $f(x)=y=9-\frac{2}{3}x$ .

The slope of the line is: $m_{1}=-\frac{2}{3}$

A perpendicular line passes through the origin to the line f (x).

The slope of this perpendicular line is: $m_{2}=-\frac{1}{m_{1}}=\frac{3}{2}$

The equation of perpendicular line passing through origin is:

$y=\frac{3}{2}x$

Compute the intersecting point between the lines as follows:

$y=9-\frac{2}{3}x\\\\\frac{3}{2}x=9-\frac{2}{3}x\\\\\frac{3}{2}x+\frac{2}{3}x=9\\\\\frac{13}{6}x=9\\\\x=\frac{54}{13}$

The value of <em>y</em> is:

$y=\frac{3}{2}x=\frac{3}{2}\times\frac{54}{13}=\frac{81}{13}$

The intersecting point is $(\frac{54}{13},\ \frac{81}{13})$ .

The <em>y</em>-intercept of the line f (x) is, 9, i.e. the point is (0, 9).

So, the triangle is bounded by the points:

(0, 0), (0, 9) and $(\frac{54}{13},\ \frac{81}{13})$

Consider the diagram attached.

Compute the area of the triangle as follows:

$\text{Area}=\frac{1}{2}\times 9\times \frac{54}{13}=18.69231\approx 18.70$

Thus, the area of the triangle is 18.70 sq.units.

7 0

4 years ago

The winners of a carnival game draw a ticket from a box to determine their prize. Each winner draws a ticket and places it back

swat32

Answer:

He generates a random whole number from 1 to 3.

Step-by-step explanation:

Each winner has 3 possibilities, gets cotton candy, gets a shirt, or gets a keychain. We can asign each of this possibiltiies a number, 1 for cotton candy, 2 for the shirt and 3 for a keychain. So, for a simulation, its enough to take one of this whole numbers for each winner.

6 0

3 years ago