Please help solve this question ?
2 Answers
See below.
Explanation:
(a). From diagram we can see area of the base is:
b). From diagram we can see that the surface areas are:
c). We now have formulas for volume and surface area:
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
Put into surface area function:
We now find:
To find max/min points we need to solve:
So:
We know for are minimum point the second derivative is positive, so we differentiate the first derivative:
Now we have the second derivative, we test the value of
This is positive, so surface area is a minimum for given volume when:
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"#