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=6-2.5H

H=X

L=1.5X

V(X)=(1.5X)(X)(6-2.5X)

V(X)=9X^2-3.73X^3

V1(X)=18X-11.25X^2

X=1.6, (X=0)

V2(X)=18-2.25X

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=6-2.5x and L=1.5x

v(x)=L*H*W= x(1.5x)(6-2.5x)= 9x^2-3.75x^3

V'(x)= 18x-11.25x^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)=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
Thank you

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