Question

alberta is designing a container in the shape of a rectangular prism to ship electronic devices. the length of the container is 10 inches longer than the height. the sum of the length, width, and height is 25 inches. write a function for the volume of the prism. what do the x - intercepts of the graph mean in this context? what dimensions of the container will maximize the volume?
0.75 - 10
(simplify your answer. use a comma to separate answers as needed.)
what do the x - intercepts of the graph mean in this context?
the intercepts 0 and 7.5 represent the heights that will result in a container with zero volume. the intercept - 10 is not meaningful because it is not possible to have a height that is negative.
what dimensions of the container will maximize the volume?
the function has a maximum when x is approximately . so the container has a maximum volume when height is about in., length is about in., and width is about in.
(type whole numbers.)

Explanation:

Step1: Define variables

Let the height of the rectangular - prism be $x$ inches. Then the length is $x + 10$ inches. Let the width be $y$ inches. Given that $x+(x + 10)+y=25$, so $y=25-(2x + 10)=15 - 2x$.

Step2: Write the volume function

The volume $V$ of a rectangular - prism is $V=\text{length}\times\text{width}\times\text{height}$. Substituting the values, we get $V(x)=(x + 10)(15 - 2x)x=x(15x-2x^{2}+150 - 20x)=-2x^{3}-5x^{2}+150x$.

Step3: Find the x - intercepts

Set $V(x)=0$, so $-2x^{3}-5x^{2}+150x = 0$. Factor out $x$: $x(-2x^{2}-5x + 150)=0$. Using the quadratic formula for $-2x^{2}-5x + 150 = 0$ ($x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a=-2$, $b=-5$, $c = 150$), we also have the solution $x = 0$ from the factored - out $x$. The non - zero root of $-2x^{2}-5x + 150 = 0$ is $x=\frac{5\pm\sqrt{25+1200}}{-4}=\frac{5\pm\sqrt{1225}}{-4}=\frac{5\pm35}{-4}$. We get $x = 7.5$ and $x=-10$. In the context of the problem, $x = 0$ and $x = 7.5$ represent the heights that result in a container with zero volume. $x=-10$ is not meaningful since height cannot be negative.

Step4: Find the maximum of the volume function

Take the derivative of $V(x)$: $V^\prime(x)=-6x^{2}-10x + 150$. Set $V^\prime(x)=0$, then $3x^{2}+5x - 75 = 0$. Using the quadratic formula $x=\frac{-5\pm\sqrt{25+900}}{6}=\frac{-5\pm\sqrt{925}}{6}=\frac{-5\pm5\sqrt{37}}{6}$. We take the positive root $x\approx3.9$.
When $x\approx4$ (rounding to the nearest whole number), the length $x + 10\approx14$ inches, and the width $15-2x\approx15 - 8 = 7$ inches.

Answer:

The function for the volume is $V(x)=-2x^{3}-5x^{2}+150x$. The $x$-intercepts $0$ and $7.5$ represent the heights that result in a container with zero volume. The container has a maximum volume when height is about $4$ in, length is about $14$ in, and width is about $7$ in.