Question

question 5 of 5
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a company that manufactures vehicle trailers estimates that the monthly profit for selling its midsize trailer is represented by function p, where t is the number of trailers sold.
$p(t) = -25t^3 + 625t^2 - 2,500t$
use the key features of function p to complete these statements.
the company makes a profit when it sells $\boldsymbol{
abla}$ trailers.
the maximum profit of approximately $ $\boldsymbol{
abla}$ occurs when it sells approximately $\boldsymbol{
abla}$ trailers.

Explanation:

Step1: Find profit-positive t values

We need to solve $p(t) = -25t^3 + 625t^2 - 2500t > 0$.
First factor the function:
$p(t) = -25t(t^2 - 25t + 100) = -25t(t-5)(t-20)$
Find roots by setting $p(t)=0$: $t=0$, $t=5$, $t=20$.
Test intervals:

• For $0
• For $5 0$ (profit)
• For $t>20$: $p(t) < 0$ (loss)

Step2: Find maximum profit (critical points)

Take derivative of $p(t)$:
$p'(t) = -75t^2 + 1250t - 2500$
Set $p'(t)=0$:
$-75t^2 + 1250t - 2500 = 0$
Divide by $-25$: $3t^2 - 50t + 100 = 0$
Use quadratic formula $t = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $a=3$, $b=-50$, $c=100$:
$t = \frac{50\pm\sqrt{2500-1200}}{6} = \frac{50\pm\sqrt{1300}}{6} = \frac{50\pm10\sqrt{13}}{6} = \frac{25\pm5\sqrt{13}}{3}$
Calculate values:
$t_1 = \frac{25+5\sqrt{13}}{3} \approx \frac{25+18.03}{3} \approx 14.34$
$t_2 = \frac{25-5\sqrt{13}}{3} \approx \frac{25-18.03}{3} \approx 2.32$
Test critical points: $t\approx2.32$ is a local minimum (loss), $t\approx14.34$ is a local maximum.

Step3: Calculate maximum profit

Substitute $t\approx14.34$ into $p(t)$:
$p(14.34) = -25(14.34)^3 + 625(14.34)^2 - 2500(14.34)$
Calculate each term:
$(14.34)^3 \approx 2954$, so $-25*2954 \approx -73850$
$(14.34)^2 \approx 205.6$, so $625*205.6 \approx 128500$
$2500*14.34 \approx 35850$
$p(14.34) \approx -73850 + 128500 - 35850 = 18800$

Answer:

The company makes a profit when it sells between 5 and 20 trailers.
The maximum profit of approximately $\$18800$ occurs when it sells approximately 14 trailers.