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, and then define the other dimensions in terms of x.
+128485 Sorry, Hayley........but this one is a Calculus problem......it's a good one, but a little tricky.....
L + W + H = 6
L = 1.5 H
Let x = H then L = 1.5 x and W = 6 - L - H = 6 - 1.5x - x = 6 - 2.5x
So.......the volume - V - is given by
V = H * L * W
V = x *( 1.5)x * [ 6 - 2.5x ] =
V = 1.5x^2 *[ 6 - 2.5x] =
V = 9x^2 - 3.75x^3 take the derivative and set to 0
V ' = 18x - 11.25x^2 = 0 factor
x [ 18 - 11.25x] = 0
So either x = 0 [reject]
Or
18 - 11.25x = 0
11.25x = 18 divide both sides by 11.25
x = 18 / 11.25 = 1.6 units........this is the height
1.5x = 1.5(1.5)= L = 2.4 units
And the width, W, = 6 - 2.5x = 6 - 2.5(1.6) = 2 units
Here's the graph of the function in terms of x ........https://www.desmos.com/calculator/fhmdwdkprw
Note that a max volume of 7.7 cubic units occurs when x = 1.6
+8581 Let the perimeter equal P
P = 2l+2w
6 = 2*(1.5x)+2x
6 = 3x+2x
6 = 5x
x = 6/5
x = 1.2 inches
Width = 1.2 inches
Height = 1.5*width = 1.8 inches
+128485 Sorry, Hayley........but this one is a Calculus problem......it's a good one, but a little tricky.....
L + W + H = 6
L = 1.5 H
Let x = H then L = 1.5 x and W = 6 - L - H = 6 - 1.5x - x = 6 - 2.5x
So.......the volume - V - is given by
V = H * L * W
V = x *( 1.5)x * [ 6 - 2.5x ] =
V = 1.5x^2 *[ 6 - 2.5x] =
V = 9x^2 - 3.75x^3 take the derivative and set to 0
V ' = 18x - 11.25x^2 = 0 factor
x [ 18 - 11.25x] = 0
So either x = 0 [reject]
Or
18 - 11.25x = 0
11.25x = 18 divide both sides by 11.25
x = 18 / 11.25 = 1.6 units........this is the height
1.5x = 1.5(1.5)= L = 2.4 units
And the width, W, = 6 - 2.5x = 6 - 2.5(1.6) = 2 units
Here's the graph of the function in terms of x ........https://www.desmos.com/calculator/fhmdwdkprw
Note that a max volume of 7.7 cubic units occurs when x = 1.6
