3. if 12 ft² of material is available to make a box with square base and open top, find the largest possible…

Explanation:

Step1: Define variables

Let the side length of the square base be (x) (in feet) and the height of the box be (y) (in feet). The surface - area formula for the open - top box is (S = x^{2}+4xy). Given (S = 12), so (12=x^{2}+4xy), and we can express (y=\frac{12 - x^{2}}{4x}). The volume formula for the box is (V=x^{2}y).

Step2: Substitute (y) into the volume formula

Substitute (y=\frac{12 - x^{2}}{4x}) into (V=x^{2}y). Then (V(x)=x^{2}\cdot\frac{12 - x^{2}}{4x}=\frac{12x - x^{3}}{4}=3x-\frac{1}{4}x^{3}), where (x>0).

Step3: Find the first derivative

Differentiate (V(x)) with respect to (x). Using the power rule ((x^{n})^\prime=nx^{n - 1}), we have (V^\prime(x)=3-\frac{3}{4}x^{2}).

Step4: Find the critical points

Set (V^\prime(x) = 0). Then (3-\frac{3}{4}x^{2}=0). [ \begin{align*} \frac{3}{4}x^{2}&=3\ x^{2}&=4\ x& = 2\quad(x>0) \end{align*} ]

Step5: Find the second derivative

Differentiate (V^\prime(x)) with respect to (x). (V^{\prime\prime}(x)=-\frac{3}{2}x). When (x = 2), (V^{\prime\prime}(2)=-\frac{3}{2}\times2=-3<0). So (V(x)) has a local maximum at (x = 2).

Step6: Find the value of (y) and (V)

When (x = 2), (y=\frac{12-2^{2}}{4\times2}=\frac{12 - 4}{8}=1). The volume (V=x^{2}y=(2)^{2}\times1 = 4) (in (ft^{3})).

Step7: Justify the absolute maximum

Since (x>0), and as (x\rightarrow0^{+}), (V(x)=3x-\frac{1}{4}x^{3}\rightarrow0); as (x\rightarrow\sqrt{12}) (when (y = 0)), (V(x)=3x-\frac{1}{4}x^{3}\rightarrow0). And we have found a local maximum at (x = 2) (using the second - derivative test (V^{\prime\prime}(2)<0)), so this local maximum is also an absolute maximum.

Answer:

The largest possible volume of the box is (4\space ft^{3}).