Question

The diagram shows an open container constructed out of 200cm^2 of cardboard. The two vertical end pieces are isosceles triangles with sides 5x cm, 5x cm, and 8x cm, and the two side pieces are rectangles of length y cm and width 5x cm. The open top is a horizontal rectangle.

Answer 1

Answer:

(i)  See below for proof.

(ii)  See below for proof.

(iii)  x = 5/3

(iv)  Maximum

Step-by-step explanation:

Question (i)

The given container is made up of the following shapes:

• Two congruent isosceles triangles, each with a base of 8x cm and congruent sides of 5x cm.
• Two congruent rectangles, each with a width of 5x cm and a length of y cm.

The formula for the area of an isosceles triangle given its base (b) and congruent side length (s) is:

$A=\dfrac{1}{2}b\sqrt{s^2-\dfrac{b^2}{4}}$

Therefore, given that b = 8x cm and s = 5x cm, the area of one isosceles triangle of the given container is:

$A_{\triangle}=\dfrac{1}{2}(8x)\sqrt{(5x)^2-\dfrac{(8x)^2}{4}}\\\\\\A_{\triangle}=4x\sqrt{25x^2-\dfrac{64x^2}{4}}\\\\\\A_{\triangle}=4x\sqrt{25x^2-16x^2}\\\\\\A_{\triangle}=4x\sqrt{9x^2}\\\\\\A_{\triangle}=4x\cdot 3x\\\\\\A_{\triangle}=12x^2\; \sf cm^2$

The area of a rectangle is the product of its width and length. Therefore, the area of one rectangle of the given container is:

$A_{\square}=5x \cdot y\\\\\\A_{\square}=5xy$

So, the equation for the total surface area of the container is:

$\textsf{Surface area of container}=2 \cdot A_{\triangle}+2 \cdot A_{\square}\\\\\\\textsf{Surface area of container}=2 \cdot 12x^2+2 \cdot 5xy\\\\\\\textsf{Surface area of container}=24x^2+10xy$

Given that the total surface area of the container is 200 cm², then:

$24x^2+10xy=200$

Solve for y:

$y=200-24x^2\\\\\\y=\dfrac{200-24x^2}{10x}$

Therefore, we have shown that:

$\Large\boxed{\boxed{y=\dfrac{200-24x^2}{10x}}}$

$\dotfill$

Question (ii)

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:

$V=12x^2 \cdot y$

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:

$V=12x^2 \cdot \dfrac{200-24x^2}{10x}\\\\\\\\V= \dfrac{12x^2(200-24x^2)}{10x}\\\\\\\\V= \dfrac{12x(200-24x^2)}{10}\\\\\\\\V= \dfrac{2400x-288x^3}{10}\\\\\\\\V=240x-28.8x^3$

Therefore, we have shown that:

$\Large\boxed{\boxed{V=240x-28.8x^3}}$

$\dotfill$

Question (iii)

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:

$V'(x)=240-3 \cdot 28.8x^{3-1}\\\\\\V'(x)=240-86.4x^2$

Now, set it to zero and solve for x:

$240-86.4x^2=0\\\\\\86.4x^2=240\\\\\\x^2=\dfrac{240}{86.4}\\\\\\x^2=\dfrac{25}{9}\\\\\\x=\sqrt{\dfrac{25}{9}}\\\\\\x=\pm\dfrac{5}{3}$

As length is positive, the value of x for which V has a stationary value is:

$\Large\boxed{\boxed{x=\dfrac{5}{3}}}$

$\dotfill$

Question (iv)

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):

$V''(x)=0-2 \cdot 86.4x^{2-1}\\\\\\V''(x)=-172.8x$

Substitute x = 5/3 into V''(x):

$V''\left(\dfrac{5}{3}\right)=-172.8\left(\dfrac{5}{3}\right)\\\\\\V''\left(\dfrac{5}{3}\right)=-288$

As V''(5/3) is negative, the stationary value is a:

$\Large\boxed{\boxed{\sf Maximum\;value}}$

Answer 2

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.

Depth

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

Area

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

(i) Equation for y

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

(ii) Volume

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 ...

  $V=12x^2\left(\dfrac{200-24x^2}{10x}\right)=1.2x(200 -24x^2)\\\\\\\boxed{V=240x-28.8x^3}$

(iii) Stationary value

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.

(iv) Maximum

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.