Answer:
(i) See below for proof.
(ii) See below for proof.
(iii) x = 5/3
(iv) Maximum
Step-by-step explanation:
The given container is made up of the following shapes:
The formula for the area of an isosceles triangle given its base (b) and congruent side length (s) is:
[tex]A=\dfrac{1}{2}b\sqrt{s^2-\dfrac{b^2}{4}}[/tex]
Therefore, given that b = 8x cm and s = 5x cm, the area of one isosceles triangle of the given container is:
[tex]A_{\triangle}=\dfrac{1}{2}(8x)\sqrt{(5x)^2-\dfrac{(8x)^2}{4}}\\\\\\A_{\triangle}=4x\sqrt{25x^2-\dfrac{64x^2}{4}}\\\\\\A_{\triangle}=4x\sqrt{25x^2-16x^2}\\\\\\A_{\triangle}=4x\sqrt{9x^2}\\\\\\A_{\triangle}=4x\cdot 3x\\\\\\A_{\triangle}=12x^2\; \sf cm^2[/tex]
The area of a rectangle is the product of its width and length. Therefore, the area of one rectangle of the given container is:
[tex]A_{\square}=5x \cdot y\\\\\\A_{\square}=5xy[/tex]
So, the equation for the total surface area of the container is:
[tex]\textsf{Surface area of container}=2 \cdot A_{\triangle}+2 \cdot A_{\square}\\\\\\\textsf{Surface area of container}=2 \cdot 12x^2+2 \cdot 5xy\\\\\\\textsf{Surface area of container}=24x^2+10xy[/tex]
Given that the total surface area of the container is 200 cm², then:
[tex]24x^2+10xy=200[/tex]
Solve for y:
[tex]y=200-24x^2\\\\\\y=\dfrac{200-24x^2}{10x}[/tex]
Therefore, we have shown that:
[tex]\Large\boxed{\boxed{y=\dfrac{200-24x^2}{10x}}}[/tex]
[tex]\dotfill[/tex]
The volume of a triangular prism is the area of the triangular base multiplied by the height of the prism.
In this case, the area of the triangular base is 12x² cm², and the height is y cm, so:
[tex]V=12x^2 \cdot y[/tex]
Plug in the expression for y found in part (i) to create an equation for the volume of the container in terms of x only:
[tex]V=12x^2 \cdot \dfrac{200-24x^2}{10x}\\\\\\\\V= \dfrac{12x^2(200-24x^2)}{10x}\\\\\\\\V= \dfrac{12x(200-24x^2)}{10}\\\\\\\\V= \dfrac{2400x-288x^3}{10}\\\\\\\\V=240x-28.8x^3[/tex]
Therefore, we have shown that:
[tex]\Large\boxed{\boxed{V=240x-28.8x^3}}[/tex]
[tex]\dotfill[/tex]
The value of x for which volume V has a stationary value is the value of x when V'(x) = 0.
Differentiate the volume equation:
[tex]V'(x)=240-3 \cdot 28.8x^{3-1}\\\\\\V'(x)=240-86.4x^2[/tex]
Now, set it to zero and solve for x:
[tex]240-86.4x^2=0\\\\\\86.4x^2=240\\\\\\x^2=\dfrac{240}{86.4}\\\\\\x^2=\dfrac{25}{9}\\\\\\x=\sqrt{\dfrac{25}{9}}\\\\\\x=\pm\dfrac{5}{3}[/tex]
As length is positive, the value of x for which V has a stationary value is:
[tex]\Large\boxed{\boxed{x=\dfrac{5}{3}}}[/tex]
[tex]\dotfill[/tex]
To determine whether x = 5/3 is a maximum or minimum stationary value, we need to substitute x = 5/3 into the second derivative V''(x). If the result is negative, it is a maximum value, and if the result is positive, it is a minimum value.
Differentiate V'(x) to find V''(x):
[tex]V''(x)=0-2 \cdot 86.4x^{2-1}\\\\\\V''(x)=-172.8x[/tex]
Substitute x = 5/3 into V''(x):
[tex]V''\left(\dfrac{5}{3}\right)=-172.8\left(\dfrac{5}{3}\right)\\\\\\V''\left(\dfrac{5}{3}\right)=-288[/tex]
As V''(5/3) is negative, the stationary value is a:
[tex]\Large\boxed{\boxed{\sf Maximum\;value}}[/tex]
Answer:
(i) y = (200 -24x²)/(10x)
(ii) V = 240x -28.8x³
(iii) x = 5/3
(iv) Maximum
Step-by-step explanation:
You want various relations between dimensions of an open triangular prism with a surface area of 200 cm², base edge lengths of 5x, 5x, and 8x, and height y.
The altitude of the isosceles triangular base can be found by considering the right triangles it creates. Each has a hypotenuse of 5x and a leg that is 8x/2 = 4x. The length of the other leg can be found from the Pythagorean theorem:
h² +(4x)² = (5x)²
h = x√(25 -16) = 3x
The area is the sum of the areas of the two triangular ends and the two rectangular sides:
A = 2(1/2bh +LW) = 2(1/2(8x)(3x) +y(5x)) = 24x² +10xy
Given the area is 200 cm², we can solve the above area equation for y:
200 = 24x² +10xy
200 -24x² = 10xy . . . . . . . subtract the term not containing y
y = (200 -24x²)/(10x) . . . . . . divide by the coefficient of y
The volume is half the product of length, width, and depth:
V = 1/2(y)(8x)(3x) = 12x²y
Using the above expression for y, this is ...
[tex]V=12x^2\left(\dfrac{200-24x^2}{10x}\right)=1.2x(200 -24x^2)\\\\\\\boxed{V=240x-28.8x^3}[/tex]
The volume will have a stationary value where its derivative with respect to x is zero.
V' = 240 -3(28.8x²) = 0
28.8x² = 80 . . . . . . . . . . . divide by 3, add 28.8x²
x² = 80/28.8 = 25/9
x = ±√(25/9) = ±5/3 . . . . . . . . . the negative value is extraneous
The value of x that makes V stationary is x = 5/3.
The cubic relation for V has a negative leading coefficient. That tells us the left-most stationary value (at x=-5/3) is a minimum, and the right-most stationary value is a maximum. We also note that as x increases through the stationary point, the sign of V' goes from positive to negative, indicating a local maximum.
The value of V at x = 5/3 is a maximum stationary value.