[latex]f(x) = 2x^3 - 15x^2 + 24x - 7[/latex] has critical numbers [latex]x =[/latex] 1 and 4. Use the Second Derivative Test for Extremes to determine whether [latex]f(1)[/latex] and [latex]f(4)[/latex] are maximums or minimums or neither.

Answer: We need to find the second derivative:

[latex]\begin{align*} f(x) & = 2x^3 - 15x^2 + 24x - 7\\ f'(x) & = 6x^2 - 30x + 24\\ f''(x) & = 12x - 30 \end{align*}[/latex]

Then we just need to evaluate [latex]f''[/latex] at each critical number:

[latex]x = 1: f''(1)=12(1)-30 \lt 0[/latex], so there is a local maximum at [latex]x = 1[/latex].

[latex]x = 4: f''(4)=12(4)-30 \gt 0[/latex], so there is a local minimum at [latex]x = 4[/latex].

The manager of a garden store wants to build a 600 square foot rectangular enclosure on the store’s parking lot in order to display some equipment. Three sides of the enclosure will be built of redwood fencing, at a cost of $7 per running foot. The fourth side will be built of cement blocks, at a cost of $14 per running foot. Find the dimensions of the least costly such enclosure.

Answer: dimensions of enclosure?

Objective: minimize cost

Objective function: [latex]C=14x+7y+7x+7y=21x+14y[/latex]

We must rewrite the objective function as a function of one variable, i.e., eliminate one of the two variables.

Since we have a condition on the area size, we have [latex]A = xy = 600[/latex] where [latex]A[/latex] is the area. Solving for [latex]x[/latex] we get [latex]x=\frac{600}{y}[/latex]. We then substitute the value of [latex]x[/latex] into [latex]C[/latex]:

[latex]C(y)=21\left(\frac{600}{y}\right)+14y=\frac{12600}{y}+14y=12600y^{-1}+14y[/latex]

Now we have a function of just one variable, so we can find the minimum, if it exists, using calculus. For that we need to find the potential locations for the minimum, i.e, we need to identify the critical numbers of [latex]C(y)[/latex].

First, the domain of [latex]C[/latex]:

mathematically: [latex]C(y)=12600y^{-1}+14y[/latex] is defined everywhere except when [latex]y=0[/latex]

contextually: [latex]y[/latex] represents the length of the enclosure, which must be a positive value.

So [latex]y>0[/latex] and the domain is [latex](0,\infty)[/latex], an open interval.

In addition, the objective function is continuous everywhere on its domain.

Then we need to find the critical numbers, which will give us the potential local minima or maxima for the cost. So we need to find out where [latex]C'(y)[/latex] is zero or undefined. Since

[latex]C'(y)=12600(-1)y^{-1-1}+14=-12600y^{-2}+14=-\frac{12600}{y^2}+14[/latex]

we have that [latex]C'(y)[/latex] is undefined for [latex]y = 0[/latex], which is not in the domain and so [latex]y=0[/latex] is not a critical number. We also have that [latex]C' (y)= 0[/latex] when

[latex]\begin{align*} -\frac{12600}{y^2}+14&=0\\\\ \Rightarrow\frac{-12600+14y^2}{y^2}&=0\\\\ \Rightarrow-12600+14y^2&=0\\\\ \Rightarrow y^2&=\frac{12600}{14}=900\\\\ \Rightarrow y&=\pm 30 \end{align*}[/latex]

Since only [latex]y=30[/latex] is in the domain, it is the only critical number.  Therefore [latex]y=30[/latex] gives the only potential maximum or minimum cost.

We now use the second derivative test to determine the answer, if possible:

[latex]C''(y)=-12600(-2)y^{-2-1}+0=25200y^{-3}=\frac{25200}{y^3} \Rightarrow C''(30)\gt 0[/latex]

so [latex]y=30[/latex] gives a local minimum.

Since this is the only critical point in the domain, which is an open interval, this must give the global minimum for the cost.

Going back to our constraint function, we can find that when [latex]y = 30[/latex], [latex]x = \frac{600}{30}=20[/latex].

Therefore, the dimensions of the enclosure that minimize the cost are 20 feet by 30 feet, where the more expensive side is 20 feet long.