Verified answer

Answer:

[tex]\qquad\boxed{ \sf{ \: \bf \: Minimum \: value \: of \: y \: is \: - 1296 \: }}\\ \\ [/tex]

Step-by-step explanation:

Given function is

[tex]\sf \: y=(17-x)(17+x)(19-x)(19+x) \\ \\ [/tex]

[tex]\sf \: y = ( {17}^{2} - {x}^{2})( {19}^{2} - {x}^{2}) \\ \\ [/tex]

[tex]\sf \: y = (289 - {x}^{2})(361 - {x}^{2}) \\ \\ [/tex]

[tex]\sf \: y = 104329 - 289 {x}^{2} - 361 {x}^{2} + {x}^{4} \\ \\ [/tex]

[tex]\sf \: y = 104329 - 650 {x}^{2} + {x}^{4} \\ \\ [/tex]

On differentiating both sides w. r. t. x, we get

[tex]\sf \: \dfrac{dy}{dx}= - 1300x + 4 {x}^{3} \\ \\ [/tex]

For maxima or minima, we have

[tex]\sf \: \dfrac{dy}{dx}= 0 \\ \\ [/tex]

[tex]\sf \: - 1300x + {4x}^{3} = 0 \\ \\ [/tex]

[tex]\sf \: 4x( - 325 + {x}^{2}) = 0 \\ \\ [/tex]

[tex]\sf\implies x = 0 \: \: or \: \: x \: = \: \pm \: \sqrt{325} \\ \\ [/tex]

Now, from above

[tex]\sf \: \dfrac{dy}{dx}= - 1300x + 4 {x}^{3} \\ \\ [/tex]

On differentiating both sides w. r. t. x, we get

[tex]\sf \: \dfrac{ {d}^{2} y}{d {x}^{2} }= - 1300+ 12 {x}^{2} \\ \\ [/tex]

Now,

[tex]\sf \: \left(\dfrac{ {d}^{2} y}{d {x}^{2} }\right)_{x = 0}= - 1300 < 0 \\ \\ [/tex]

[tex]\bf\implies \: \: y \: is \: maximum \\ \\ [/tex]

Now,

[tex]\sf \: \left(\dfrac{ {d}^{2} y}{d {x}^{2} }\right)_{x = \sqrt{325} }= - 1300 + 12 \times 325 = - 1300 + 3900 = 2600 > 0 \\ \\ [/tex]

[tex]\bf\implies \: \: y \: is \: minimum \: at \: x = \sqrt{325} \\ \\ [/tex]

Again Consider,

[tex]\sf \: \left(\dfrac{ {d}^{2} y}{d {x}^{2} }\right)_{x = - \sqrt{325} }= - 1300 + 12 \times 325 = - 1300 + 3900 = 2600 > 0 \\ \\ [/tex]

[tex]\bf\implies \: \: y \: is \: minimum \: at \: x = - \sqrt{325} \\ \\ [/tex]

Now,

[tex]\sf \: Minimum \: value \: at \: x = \sqrt{325} \: is \: \\ \\ [/tex]

[tex]\qquad\sf \: = \: (289 - 325)(361 - 325) \\ \\ [/tex]

[tex]\qquad\sf \: = \: ( - 36)(36) \\ \\ [/tex]

[tex]\qquad\sf \: = \: - 1296 \\ \\ [/tex]

Now,

[tex]\sf \: Minimum \: value \: at \: x = - \sqrt{325} \: is \: \\ \\ [/tex]

[tex]\qquad\sf \: = \: (289 - 325)(361 - 325) \\ \\ [/tex]

[tex]\qquad\sf \: = \: ( - 36)(36) \\ \\ [/tex]

[tex]\qquad\sf \: = \: - 1296 \\ \\ [/tex]

Hence,

[tex]\sf\implies \bf \: Minimum \: value \: of \: y \: is \: - 1296 \\ \\ [/tex]

Answer:

Answer:

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−

1296

Minimumvalueofyis−1296

Step-by-step explanation:

Given function is

�

=

(

17

−

�

)

(

17

+

�

)

(

19

−

�

)

(

19

+

�

)

y=(17−x)(17+x)(19−x)(19+x)

�

=

(

17

2

−

�

2

)

(

19

2

−

�

2

)

y=(17

2

−x

2

)(19

2

−x

2

)

�

=

(

289

−

�

2

)

(

361

−

�

2

)

y=(289−x

2

)(361−x

2

)

�

=

104329

−

289

�

2

−

361

�

2

+

�

4

y=104329−289x

2

−361x

2

+x

4

�

=

104329

−

650

�

2

+

�

4

y=104329−650x

2

+x

4

On differentiating both sides w. r. t. x, we get

�

�

�

�

=

−

1300

�

+

4

�

3

dx

dy

=−1300x+4x

3

For maxima or minima, we have

�

�

�

�

=

0

dx

dy

=0

−

1300

�

+

4

�

3

=

0

−1300x+4x

3

=0

4

�

(

−

325

+

�

2

)

=

0

4x(−325+x

2

)=0

⟹

�

=

0

�

�

�

=

±

325

⟹x=0orx=±

325

Now, from above

�

�

�

�

=

−

1300

�

+

4

�

3

dx

dy

=−1300x+4x

3

On differentiating both sides w. r. t. x, we get

�

2

�

�

�

2

=

−

1300

+

12

�

2

dx

2

d

2

y

=−1300+12x

2

Now,

(

�

2

�

�

�

2

)

�

=

0

=

−

1300

<

0

(

dx

2

d

2

y

)

x=0

=−1300<0

⟹

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⟹yismaximum

Now,

(

�

2

�

�

�

2

)

�

=

325

=

−

1300

+

12

×

325

=

−

1300

+

3900

=

2600

>

0

(

dx

2

d

2

y

)

x=

325

=−1300+12×325=−1300+3900=2600>0

⟹

�

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�

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=

325

⟹yisminimumatx=

325

Again Consider,

(

�

2

�

�

�

2

)

�

=

−

325

=

−

1300

+

12

×

325

=

−

1300

+

3900

=

2600

>

0

(

dx

2

d

2

y

)

x=−

325

=−1300+12×325=−1300+3900=2600>0

⟹

�

�

�

�

�

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=

−

325

⟹yisminimumatx=−

325

Now,

�

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=

325

�

�

Minimumvalueatx=

325

is

=

(

289

−

325

)

(

361

−

325

)

=(289−325)(361−325)

=

(

−

36

)

(

36

)

=(−36)(36)

=

−

1296

=−1296

Now,

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

=

−

325

�

�

Minimumvalueatx=−

325

is

=

(

289

−

325

)

(

361

−

325

)

=(289−325)(361−325)

=

(

−

36

)

(

36

)

=(−36)(36)

=

−

1296

=−1296

Hence,

⟹

�

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�

−

1296

⟹Minimumvalueofyis−1296