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
- algebra - kay, Monday, November 26, 2012 at 10:39am
If H=x then W=6-2.5x and L=1.5x
v(x)=L*H*W= x(1.5x)(6-2.5x)= 9x^2-3.75x^3
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)=18-22.5x 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^2-3.73X^3 you've made a mistake as it's 3.75 not 3.73
- algebra - lee, Monday, November 26, 2012 at 7:07pm
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