Question

part a: written answer
instructions:
please write your solution on the written answer sheet template provided.
you do not need to rewrite the question prompt on your submission.
please answer each question fully, providing all reasoning. you will be graded based on both mathematical correctness and clarity of writing according to the written portion rubric.
question:
a closed rectangular container with a square base is to be constructed for a total cost of $48. the material for the bottom and the lid of the box costs $2/cm² and the material for the sides costs $1/cm². your goal is to find the dimensions for the box that maximizes its volume.
(a) write the volume of the box as a function of a single variable, and state the domain of this function. clearly define any notation or variables you introduce.
(b) find the dimensions (base length and height) of the box that maximizes its volume. fully justify your answer with appropriate mathematical reasoning.
hints

• the volume of a box is equal to the area of the base multiplied by the height of the box.
• justify your result using an appropriate method like first derivative test, second derivative test, closed interval method, or global interval method. state which method youre using and verify all required conditions. (not all methods listed may be suitable for this problem.)

please write your solution on the written answer sheet template provided.

Explanation:

Step1: Define variables

Let the side length of the square base be \(x\) (in cm) and the height of the box be \(h\) (in cm).
The cost function \(C = 2(2x^{2})+4(xh)\) (since bottom and lid cost \(\$2/cm^{2}\) and sides cost \(\$1/cm^{2}\), and \(C = 48\)). So \(4x^{2}+4xh=48\), which simplifies to \(h=\frac{12 - x^{2}}{x}\).
The volume function \(V=x^{2}h\). Substitute \(h\) into \(V\): \(V(x)=x^{2}\cdot\frac{12 - x^{2}}{x}=12x - x^{3}\).
For the domain, \(x>0\) (length cannot be negative) and \(h=\frac{12 - x^{2}}{x}>0\). Solving \(\frac{12 - x^{2}}{x}>0\) (since \(x>0\), we can multiply both sides by \(x\) without changing the inequality sign) gives \(12 - x^{2}>0\), so \(x^{2}<12\) or \(x < 2\sqrt{3}\). Thus, the domain is \((0,2\sqrt{3})\).

Step2: Find the derivative of \(V(x)\)

Differentiate \(V(x)=12x - x^{3}\) using the power rule \((x^{n})^\prime=nx^{n - 1}\).
\(V^\prime(x)=12-3x^{2}\).
Set \(V^\prime(x) = 0\): \(12-3x^{2}=0\).
\(3x^{2}=12\), \(x^{2} = 4\), \(x = 2\) (since \(x\in(0,2\sqrt{3})\)).

Step3: Use the Second - Derivative Test

Differentiate \(V^\prime(x)\) to get \(V^{\prime\prime}(x)=-6x\).
When \(x = 2\), \(V^{\prime\prime}(2)=- 12<0\). So \(V(x)\) has a local maximum at \(x = 2\).
When \(x = 2\), \(h=\frac{12-4}{2}=4\).

Answer:

(a) The volume function is \(V(x)=12x - x^{3}\) with domain \((0,2\sqrt{3})\).
(b) The base length \(x = 2\) cm and height \(h = 4\) cm.