Posted by kevin on Monday, November 26, 2012 at 7:17am.
I need someone to check my work not sure if I'm doing this right.
The design of a digital box camera maximizes the volume while keeping the sum of the dimensions at 6 inches. If the length must be 1.5 times the height, what should each dimension be? HINT: Let x represent one of the dimensions, then define the other dimensions in terms of x.
I got h=1.6 in
w=2in
l=2.4in
l+w=H=6
L=1.5H
1.5H+W+H=6
2.5W=6
W=62.5H
H=X
L=1.5X
V(X)=(1.5X)(X)(62.5X)
V(X)=9X^23.73X^3
V1(X)=18X11.25X^2
X=1.6, (X=0)
V2(X)=182.25X
V2(16)=18<0
X=16
H=1.6
W=2
L=2.4

algebra  kay, Monday, November 26, 2012 at 10:39am
L+W+H=6
L=1.5*H
2.5H+w=6
If H=x then W=62.5x and L=1.5x
v(x)=L*H*W= x(1.5x)(62.5x)= 9x^23.75x^3
V'(x)= 18x11.25x^2
For stationary point; V'(x)=0 so x=0 reject or 18=11.25x so x=18/11.25= 1.6
check if it's a max: V''(x)=1822.5x at x=1.6 V''(1.6)= 18<0 so yes
H=1.6, W=2, L= 2.4
YOUR ANSWER IS CORRECT. JUST IN LINE V(X)=9X^23.73X^3 you've made a mistake as it's 3.75 not 3.73

algebra  lee, Monday, November 26, 2012 at 7:07pm
Thank you
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