# (a) You have been asked to bid on the construction of a square-bottomed box with no top which...

## Question:

(a) You have been asked to bid on the construction of a square-bottomed box with no top which will hold 100 cubic inches of water. If the bottom and sides are made from the same material, what are the dimensions of the box which uses the least material? (Assume that no material is wasted.)

(b) Suppose the box in part (a) uses different materials for the bottom and the sides. If the bottom material costs 5 cents per square inch and the side material costs 3 cents per square inch, what are the dimensions of the least expensive box which will hold 100 cubic inches of water?

## Application of Derivative:

The above problem concerns the topic of the application of the derivative to determine the maxima and minima of the function. For that, first we find the critical point by equating the first derivative of the function to zero. The critical point is the point where function attains its maximum or minimum value.

The Volume of the cuboid is given as {eq}L\times W \times H {/eq} where L is length , W is width and H is height of the box.

Given the base of the box is square so let the base is having length and width {eq}x \ in {/eq} and the volume of the box must be {eq}100 \text{ inch}^3 {/eq}.

Also assume that height is {eq}h\ in {/eq}.

Now the volume of the box is given as

{eq}\displaystyle \begin{align} V &=(x)(x)(h) \\ 100 &=x^2h\\ h &=\frac{100}{x^2} \end{align} {/eq}.

a In first part, we have to minimize the material used, that means we have to minimize the surface area.

Total surface area of the box is given by

{eq}\displaystyle S=(x^2)+(4xh)\\ {/eq}

Substituting the value of h, we obtain

{eq}\displaystyle \begin{align} S &=x^2+4x\frac{100}{x^2}\\ \displaystyle S &=x^2+\frac{400}{x}\\ \end{align} {/eq}

Now we have to minimize the surface area, so we differentiate cost with respect to {eq}x {/eq}, and then equate it to zero

{eq}\displaystyle \begin{align} S &=x^2+\frac{400}{x}\\ \frac{\mathrm{d} S}{\mathrm{d} x} &=\frac{\mathrm{d} }{\mathrm{d} x}\left (x^2+\frac{400}{x} \right )\\ \frac{\mathrm{d} S}{\mathrm{d} x} &=\left (2x-\frac{400}{x^2} \right )=0\\ x^3 &=200 \\ x &=5.85 \ in \end{align} {/eq}

So the length and width of the box is {eq}{\color{Blue} {x=5.85 \ in}} {/eq}, and height is {eq}{\color{Blue} {\displaystyle h=\frac{100}{x^2}=\frac{100}{5.85^2}=2.92 \ in}} {/eq}

b In second part, we have to minimize the total cost of material used. For that first we find the cost function. Which is given as

{eq}\displaystyle C(x)=5(x^2)+3(4xh)\\ {/eq}

Substituting the value of h, we obtain

{eq}\displaystyle \begin{align} C(x) &=5x^2+12x\frac{100}{x^2}\\ \displaystyle C(x) &=5x^2+\frac{1200}{x}\\ \end{align} {/eq}

Now to minimize the cost, we differentiate cost function with respect to {eq}x {/eq}, and then equate it to zero, we obtain

{eq}\displaystyle \begin{align} C(X)&=5x^2+\frac{1200}{x}\\ \frac{\mathrm{d} C(x)}{\mathrm{d} x} &=\frac{\mathrm{d} }{\mathrm{d} x}\left (5x^2+\frac{1200}{x} \right )\\ \frac{\mathrm{d} C(x)} {\mathrm{d} x} &=\left (10x-\frac{1200}{x^2} \right )=0\\ x^3 &=120 \\ x &=4.93 \ in \end{align} {/eq}

So the length and width of the box is {eq}{\color{Blue} {x=4.93 \ in}} {/eq}, and height is {eq}{\color{Blue} {\displaystyle h=\frac{100}{x^2}=\frac{100}{4.93^2}=4.11\ in}} {/eq} 