Question

Please help solve this question ?

Answer 1

See below.

Explanation:

(a). From diagram we can see area of the base is:

`2x*x`

`h` equals height, so volume is:

`2x*x*h=2x^2h`

`color(blue)(V=2x^2h)`

b). From diagram we can see that the surface areas are:

`2*(2x*x)+2*(xh)+2*(2x*h)=4x^2+6hx`

`color(blue)(A=4x^2+6hx)`

c). We now have formulas for volume and surface area:

`V=2x^2h and A=4x^2+6xh`

We need minimum surface area for given volume. We need to combine the equation for the given volume into the equation for surface area. We are only differentiating in terms of `x` so we need to find `h` in terms of `x` using the volume equation.

`1000=2x^2h=>h=500/x^2`

Put into surface area function:

`A=4x^2+6x(500/x^2)=4x^2+3000/x`

`A=4x^2+3000/x`

We now find:

`(dA)/dx(4x^2+3000/x)`

`(dA)/dx(4x^2+3000/x)=8x-3000/x^2`

To find max/min points we need to solve:

`(dA)/dx=0`

So:

`8x-3000/x^2=0`

`8x^3-3000=0`

`8x^3-3000=3000/8=>x=root(3)(3000/8)=root(3)(3000)/2`

We know for are minimum point the second derivative is positive, so we differentiate the first derivative:

`(d^2S)/(dx^2)=(dS)/(dx)(8x-3000/x^2)=8+6000/x^3`

Now we have the second derivative, we test the value of `x` from:

`d/dx=0` which is `x=root(3)(3000)/2`

`8+6000/(root(3)(3000)/2)^3=24`

This is positive, so surface area is a minimum for given volume when:

`x=root(3)(3000)/2~~7.211color(white)(888)` 3 .d.p.

Answer 2

`"see explanation"`

Explanation:

`(a)`

`V=lbh=2x xx x xxh=2x^2h`

`"now "V=1000`

`rArr2x^2h=1000rArrh=1000/(2x^2)=500/x^2`

`(b)`

`A="front/back "+" 2 ends "+" top/bottom"`

`color(white)(A)=2(2hx)+2(hx)+2(2x^2)`

`color(white)(A)=4hx+2hx+4x^2`

`color(white)(A)=6hx+4x^2`

`"from "(a)" we have "h=500/x^2`

`rArrA=6x xx500/x^2+4x^2=3000/x+4x^2`

`(c)`

`A=3000/x+4x^2=3000x^-1+4x^2`

`rArr(dA)/(dx)=-3000x^-2+8x=-3000/x^2+8x`

`"for max/min "(dA)/(dx)=0`

`rArr-3000/x^2+8x=0`

`"multiply through by "x^2`

`rArr-3000+8x^3=0`

`rArrx^3=3000/8=375`

`rArrx=root(3)(375)~~7.211" to 3 dec. places"`

`"to test this value gives a minimum use the"`

`color(red)"second derivative test"`

`• " if "(d^2A)/(dx^2)>0" then minimum"`

`• " if "(d^2A)/(dx^2)<0" then maximum"`

`(d^2A)/(dx^2)=6000x^-3+8=6000/x^3+8`

`"for "x~~7.211to(d^2A)/(dx^2)>0" hence minimum"`

`rArrx~~7.211" for minimum area"`

`"and "h=500/(7.211)^2~~9.616" to 3 dec. places"`