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1.

Show that among all positive numbers `x`and `y`with `x^2+y^2=r^2,`the sum `x+y`is largest when `x=y=r/(sqrt(2))`.

Answer» Here, `x^2+y^2 = r^2`
`=> y = sqrt(r^2-x^2)`
Let `S = x+y`
`=> S = x+sqrt(r^2-x^2)`
For `S` to be largest, `(dS)/dx =0`.
`:. (dS)/dx = 1+1/sqrt(r^2-x^2)(-2x)`
`:. 1+1/(2sqrt(r^2-x^2))(-2x) = 0`
`=>x = sqrt(r^2-x^2)`
`=>x^2 = r^2-x^2`
`=>2x^2 = r^2`
`=>x = +-r/sqrt2`
As `x` is a positive number,
`:. x = r/sqrt2`.
Now, `y = sqrt(r^2-x^2) = sqrt(r^2-r^2/2) = r/sqrt2`
To further prove, that this is largest sum, we can take `(d^2S)/dx^2` at `x = r/sqrt2`.
It will come negative, which proves sum `S` is largest at these values.
So, `x+y` is largest when `x = y = r/sqrt2`.
Discussion
2.

The total cost of producing `x`radio sets per day is `R sdot((x^2)/4 35 x+25)`and the price per set at which they may be sold is `R sdot(50-x/2)dot`Find the daily output to maximize the total profit.

Answer» Here, total cost of `x` radios ` = x^2/4+35x+25` Rs
Selling price of `x` radios ` = x(50-x/2) = 50x-x^2/2` Rs
`:.` Profit `(P)= 50x-x^2/2 - (x^2/4+35x+25)` Rs
`=> P = -3/4x^2 + 15x -25`
`=>(dP)/dx = -3/2x + 15`
For, maximum profit, `(dP)/dx` should be `0`.
`=> -3/2x + 15 = 0`
`=> x = 10`
So, to maximize the total profit, `x` should be `10`.
Discussion
3.

Prove that the semi-vertical angle of the right circular cone of givenvolume and least curved surface is `cot^(-1)(sqrt(2))dot`

Answer» `V=1/3pir^2h`
`h=(3V)/(pir^2)`
`CSA=pirl`
`=pirsqrt(h^2+r^2)`
`=pirsqrt((9V^2+pi^2r^6)/(pir^2)`
`CSA=sqrt((9V^2+pi^2r^6)/r`
Differentiate with respect to r
`d/(dr)(CSA)=(3pi^2r^6-9V^2-pi^2r^6)/(r^2sqrt(9V^2+pi^2r^6)=0`
`2pi^2r^6-9V^2=0`
`r^6=(9V^2)/(2pi^2)`
`r=(9/2)^(1/6)*(V/pi)^(1/3)`
`cottheta=B/P=r/h=r/((3V)/(pir^2))`
`=pi/(3V)*[(n/2)^(1/6)*(V/pi)^(1/3)]^3`
`tantheta=1/sqrt2`
`cottheta=sqrt2`
``theta=cot^(-1)sqrt2`.
Discussion
4.

A straight line is drawn through a given point `P(1,4)dot`Determine the least value of the sum of the intercepts on thecoordinate axes.

Answer» `y-y_1=m(x-x_1)`
`(x_1,y_1)=(1,4)`
`y-y_!=m(x-1)`
at y intercept,x=0
`y=4-m`
at x intercept,y=0
`x-1=-4/m`
`x=(m-4)/m`
`S=((m-4)/m)+(4-m)`
`(dS)/(dm)=(0+4/m^2-1)=0`
`4/m^2=1`
`m=pm2`
`(d^2S)/(dm^2)=-8/m^3`
When m=2,`(d^2S)/(dm^2)=-8/8=-1<0`
This will be maximum
When m=-2,`(d^2S)/(dm^2)=-8/-8=1>0`
This will be minimum
`x=(m-4)/m=(-2-4)/(-2)=3`
`y=4-m=4+2=6`
S=3+6=9.
Discussion
5.

The strength of a beam varies as the product of its breadth and squareof its depth. Find the dimensions of the strongest beam which can be cut from a circular log of radius `adot`

Answer» `b^2/4+d^2/4=a^2`
`d^2=4a^2-b^2`
`(dS)/(db)=b(-2b)+(4a^2-b^2)`
`=-2b^2+4a^2-b^2`
`=4a^2-3b^2=0`
`b^2=4/3a^2`
`b=pm2/sqrt3`
`d^2=4a^2-b^2`
`=4a^2-4/3a^2`
`d=2sqrt(2/3)a`
`(d^2S)/(db^2)=-4b-2b`
`=-6b<0`
this will be maximum.
Discussion
6.

A particle is moving in a straight line such that its distance `s`at any time `t`is given by `s=(t^4)/4-2t^3+4t^2-7.`Find when its velocity is maximum and acceleration minimum.

Answer» `S=t^4/4-2t^3+4t^2-7`
`V=(dS)/(dt)=t^3-6t^2+8t`
`a=(dV)/(dt)=(d^2S)/(dt^2)=3t^2-12t+8=0`
`t=(12pmsqrt(144-96))/6`
`t=2pm2/sqrt3`
`(d^2V)/(dt^2)=6t-12`
`t=2+2/sqrt3`
`(d^2V)/(dt^2)=6(2+2/sqrt3)-12`
@` t=2-2/sqrt3`, V is minimum
`(da)/(dt)=6t-12=0`
`(d^2a)/(dt^2)=6>0`
@`t=2` this willbe minimum.
Discussion
7.

Determine the points on the curve `x^2=4y`which are nearest to the point (0,5).

Answer» `d=sqrt((x_1-o)^2+(y_1-5)^2)`
`d=sqrt(4y_1+(y_1-5)^2`
`d/(dy_1)=1/(2sqrt(4y_1+(y_1-5)^2))*(4+2(y_1-5))=0`
`4+(2(y_1-5)=0`
`y_1=3`
`x_1^2=4y_1`
`x_1=pm2sqrt3`
Points=`(2sqrt3,3),(-2sqrt3,3).`
Discussion
8.

A rectangle is inscribed in a semi-circle of radius `r`with one of its sides on diameter of semi-circle. Find the dimensionsof the rectangle so that its area is maximum. Find also the area.

Answer» `b=sqrt(r^2-(a/2)^2`
`b=sqrt(4r^2-a^2)/2`
Area of A=`a*b=a*(sqrt(4r^2-a^2)/2)`
diff with respect to a
`(dA)/(da)=sqrt(4r^2-a^2)/2`
`=a/2*1/2*1/sqrt(4r^2-a^2)*-2a`
`(dA)/(da)=sqrt(4r^2-a^2)/2-a^2/(2sqrt(4r^2-a^2))=0`
`(4r^2-a^2-a^2)/(2sqrt(4r^2-a^2))=0`
`4r^2=2a^2`
`a=pmsqrt2r`
`b=sqrt(4r^2-a^2)/2=sqrt2r/2=4/sqrt2`
Area=`r^2.`
Discussion