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mestny [16]
3 years ago
8

Clem and Clyde have a farm with three different crops: a square field of corn, a rectangular field of artichokes, and a right-tr

iangle grove of walnut trees (as shown at right). A fence totally surrounds the farm. Find the total area of Clem and Clyde's land in square miles and tell them how much fencing they need to enclose the outside of their farm.
Mathematics
1 answer:
antoniya [11.8K]3 years ago
3 0

Answer:

The total area of the land is 4.4 square mi.

The length of the fence should be 9.2 mi.

Step-by-step explanation:

Consider the provided information.

The required figure is shown below:

First, we need to find the Hypotenuse of the triangle by using the Pythagoras theorem.

(\text{Hyp})^2=(1.6)^2+(1.2)^2

(\text{Hyp})^2=2.56+1.44

(\text{Hyp})^2=4

\text{Hyp}=2

So, the hypotenuse of the triangle is 2 mi. Also, the hypotenuse of the triangle is the side of the rectangle.

Total area = Area of square + Area of Triangle + Area of rectangle

Total area = 1.2\times 1.2+\frac{1}{2}(1.2)(1.6)+2\times 1

Total area = 1.44+0.96+ 2

Total area = 4.4 square mi.

Now calculate the fencing they need to enclose the outside of their farm.

For this add the length of the red lines shown in the figure below:

\text{Fence length outside the farm}=1.2+1.6+1+2+1+1.2+1.2

\text{Fence length outside the farm}=9.2

Hence, the length of the fence should be 9.2 mi.

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A piece of cardboard is 13 inches by 26 inches. A square is to be cut from each corner and the sides folded up to make an open-t
Vanyuwa [196]

Answer:

Hence the maximum possible volume will be the 778.53 c.c

Step-by-step explanation:

Given:

A rectangle with 13 x 26 dimensions

And corners are cut to form side squares.

To Find:

Maximum possible volume for box

Solution :

Consider a rectangle of 13 x 26 dimension with and side of square  at corner be x.

(Refer the attachment)

Now,

Formulating the volume equation for the box

So corner square sides we are going to fold up which makes height of the box

and remaining part will be length and breadth

As shown in fig,

Length=26-x

breadth=13-x

And height will be x

V(x)=x*(26-x)*(13-x)

To get maximum volume differentiate the above equation,

V(x)=x*(26*13-26*x-13*x+x^2)

V(x)=x^3-39x^2+338x\\

V'(x)=3x^2-78x+338

V''(x)=6x-78

Now ,Solve the Quadratic Equation to get x values,

3x^2-78x+338=0

x=[-b±(b^2-4ac)^1/2]/2a

x=[78±Sqrt[(78)^2-4*338*3)]/2*3

x=[78±Sqrt(3028)]/6

x=[78±55.027]/6

x=78+55.027/6 or x=78-55.027/6

x=22.17  or x=3.8288

Use these values in 6x-78 to know which value posses the max and min value for the function.

So when x=22.17

6x-78=6*22.17-78

=55.02>0  i.e function will have minimum value .

When x=3.8288

6*3.8288-78

=-55.0272<0 i.e. Function will have maximum value

Now, the function will defines the maximum volume

V(x)=x^3-39x^2+338x

V(x)=3.8288^3-39*(3.82883)^2+338*3.8288

V(x)=56.13-571.73+1294.13

V(x)=778.53 C.C

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4 years ago
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