Please help with this question

Find the dimensions of the rectangle with area 256 square inches that has minimum perimeter, and then find the minimum perimeter

mezya [45]

Answer:

Dimensions: $A=a\cdot b=256$

Perimiter: $P=2a+2b$

Minimum perimeter: [16,16]

Step-by-step explanation:

This is a problem of optimization with constraints.

We can define the rectangle with two sides of size "a" and two sides of size "b".

The area of the rectangle can be defined then as:

$A=a\cdot b=256$

This is the constraint.

To simplify and as we have only one constraint and two variables, we can express a in function of b as:

$b=\frac{256}{a}$

The function we want to optimize is the diameter.

We can express the diameter as:

$P=2a+2b=2a+2*\frac{256}{a}$

To optimize we can derive the function and equal to zero.

$dP/da=2+2\cdot (-1)\cdot\frac{256}{a^2}=0\\\\\frac{512}{a^2}=2\\\\a=\sqrt{512/2}= \sqrt {256} =16\\\\b=256/a=256/16=16$

The minimum perimiter happens when both sides are of size 16 (a square).

4 0

3 years ago

I will give brainliest

jasenka [17]

Answer:

24 Units.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Ariana and tom Tabitha recorded the amount of time spent on homework on 10 school daze graph of the data are shown

Sedaia [141]

Answer:

theres no picture

Step-by-step explanation:

5 0

3 years ago

Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )

DedPeter [7]

(a) First find the intersections of $y=e^{2x-x^2}$ and $y=2$ :

$2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}$

So the area of $R$ is given by

$\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx$

If you're not familiar with the error function $\mathrm{erf}(x)$ , then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line $y=1$ with $y=e^{2x-x^2}$ .

$1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2$

So the area of $S$ is given by

$\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx$
$\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx$

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve $y=e^{2x-x^2}$ and the line $y=1$ , or $e^{2x-x^2}-1$ . The area of any such circle is $\pi$ times the square of its radius. Since the curve intersects the axis of revolution at $x=0$ and $x=2$ , the volume would be given by

$\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx$

5 0

3 years ago

The food is made up of 1 part sugar and 4 parts water. She uses 34 cup of sugar to make the hummingbird food. How much water sho

mash [69]

34*4=136 cups of water

7 0

2 years ago