Question

A storage box with a square base must have a volume of 90 cubic centimeters. the top and bottom cost $0.60 per square centimeter and the sides cost $0.30 per square centimeter. find the dimensions that will minimize cost. (let x represent the length of the sides of the square base and let y represent the height.

Answer 1

Answer:

side length of 3.56 cm and height of 7.10 cm

Step-by-step explanation:

Answer 2

Answer-

For side length of 3.56 cm and height of 7.10 cm the cost will be minimum.

Solution-

Let us assume that,

x represents the length of the sides of the square base,

y represent the height.

Given the volume of the box is 90 cm³, so

$\Rightarrow V=90\\\\\Rightarrow x^2\times y=90\\\\\Rightarrow y=\dfrac{90}{x^2}$

As the top and bottom cost $0.60 per cm² and the sides cost $0.30 per cm². Total cost C will be,

$C=\text{cost for top and bottom}+\text{cost for rest 4 sides}\\\\=(2x^2\times 0.6)+(4xy\times 0.3)\\\\=(2x^2\times 0.6)+(4x\times \frac{90}{x^2}\times 0.3)\\\\=1.2x^2+ \dfrac{108}{x}$

Then,

$C'=\dfrac{d}{dx}(1.2x^2+ \dfrac{108}{x})=2.4x-\dfrac{108}\\\\C''=\dfrac{d^2}{dx^2}(1.2x^2+ \dfrac{108}{x})=2.4+\dfrac{216}{x^3}$

As C'' has all positive terms  so, for every positive value of x (as length can't be negative), C'' is positive.

So, for minima C' = 0

$\Rightarrow 2.4x-\dfrac{108}{x^2}=0\\\\\Rightarrow 2.4x=\dfrac{108}{x^2}\\\\\Rightarrow x^3=\dfrac{108}{2.4}=45\\\\\Rightarrow x=3.56$

Then,

$y=\dfrac{90}{x^2}$

$y=\dfrac{90}{3.56^2}$

$y=7.10$

Therefore, for side length of 3.56 cm and height of 7.10 cm the cost will be minimum.