Question

A box with a square base and open top must have a volume of 157216 cm3. We wish to find the dimensions of the box that minimize the amount of material used.
First, find a formula for the surface area of the box in terms of only x, the length of one side of the square base.
[Hint: use the volume formula to express the height of the box in terms of x.]
Simplify your formula as much as possible.
A(x)=
Next, find the derivative A′(x).
A'(x)=
Now, calculate when the derivative equals zero, that is, when A′(x)=0. [Hint: multiply both sides by x2.]
A'(x)=0 when x=
We next have to make sure that this value of x gives a minimum value for the surface area. Let's use the second derivative test. Find A"(x).
A"(x)=
Evaluate A"(x) at the x-value you gave above.

Answer 1

Answer:

• Base Length of 68cm
• Height of 34 cm.

Step-by-step explanation:

Given a box with a square base and an open top which must have a volume of 157216 cubic centimetre. We want to minimize the amount of material used.

Step 1:

Let the side length of the base =x

Let the height of the box =h

Since the box has a square base

Volume $=x^2h=157216$

$h=\dfrac{157216}{x^2}$

Surface Area of the box = Base Area + Area of 4 sides

$A(x,h)=x^2+4xh\\ Substitute h=\dfrac{157216}{x^2}\\A(x)=x^2+4x\left(\dfrac{157216}{x^2}\right)\\A(x)=\dfrac{x^3+628864}{x}$

Step 2: Find the derivative of A(x)

$If\:A(x)=\dfrac{x^3+628864}{x}\\A'(x)=\dfrac{2x^3-628864}{x^2}$

Step 3: Set A'(x)=0 and solve for x

$A'(x)=\dfrac{2x^3-628864}{x^2}=0\\2x^3-628864=0\\2x^3=628864\\x^3=314432\\x=\sqrt[3]{314432}\\ x=68$

Step 4: Verify that x=68 is a minimum value

We use the second derivative test

$If\:A(x)=\dfrac{x^3+628864}{x}\\A''(x)=\dfrac{2x^3+1257728}{x^3}\\ When x=68\\A''(x)=6$

Since the second derivative is positive at x=68, then it is a minimum point.

Recall:

$h=\dfrac{157216}{x^2}\\h=\dfrac{157216}{68^2}=34$

Therefore, the dimensions that minimizes the box surface area are:

• Base Length of 68cm
• Height of 34 cm.