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Mashcka [7]
2 years ago
6

Please help, my computer is about to die and this is due in a few minutes. Will mark brainlist if correct

Mathematics
1 answer:
SpyIntel [72]2 years ago
5 0

Answer:

x = 1.5

Step-by-step explanation:

Based in the tangent-secant theorem,

KI*JI = HI²

KI = (x + 0.5)

JI = 0.5

HI = 1

Plug in the values

(x + 0.5)(0.5) = 1²

0.5x + 0.25 = 1

0.5x = 1 - 0.25

0.5x = 0.75

x = 0.75/0.5

x = 1.5

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

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Given:

A rectangle with 13 x 26 dimensions

And corners are cut to form side squares.

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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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3 years ago
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