Answer:
x = 5 ft
y = 5 ft
h = 2.5 ft
Step-by-step explanation:
The surface area of the base of the box will be at its maximum when the difference between the length and width at its minimum.
Therefore, let y = x.
Therefore, the equation for the surface area of box is:
[tex]\implies A=xy+2xh+2yh[/tex]
Substituting y = x:
[tex]\implies A=x^2+2xh+2xh[/tex]
[tex]\implies A=x^2+4xh[/tex]
Given the surface area of the box is 75 square feet:
[tex]\implies 75=x^2+4xh[/tex]
Rearrange the equation to create an expression for h in terms of x:
[tex]\implies h=\dfrac{75-x^2}{4x}[/tex]
The equation for the volume of the box is:
[tex]\implies V=xyh[/tex]
Substitute y = x and the expression for h to create an equation for volume in terms of x:
[tex]\implies V= \dfrac{x^2(75-x^2)}{4x}[/tex]
[tex]\implies V= \dfrac{x(75-x^2)}{4}[/tex]
[tex]\implies V= \dfrac{75}{4}x-\dfrac{1}{4}x^3[/tex]
To find the value of x that maximizes the volume, differentiate V with respect to x and find the value(s) of x that makes dV/dx = 0.
[tex]\implies \dfrac{\text{d}V}{\text{d}x}=\dfrac{75}{4}-\dfrac{3}{4}x^2[/tex]
Set it to zero and solve for x:
[tex]\implies \dfrac{75}{4}-\dfrac{3}{4}x^2=0[/tex]
[tex]\implies \dfrac{3}{4}x^2=\dfrac{75}{4}[/tex]
[tex]\implies 3x^2=75[/tex]
[tex]\implies x^2=25[/tex]
[tex]\implies x=5[/tex]
Check to see if this value of x gives a minimum for V by inputting it into the second derivative:
[tex]\implies \dfrac{\text{d}^2V}{\text{d}x^2}=-\dfrac{3}{2}x[/tex]
[tex]x=5\implies \dfrac{\text{d}^2V}{\text{d}x^2}=-\dfrac{3}{2}(5)=-\dfrac{15}{2} < 0 \implies \sf maximum[/tex]
Finally, substitute the found value of x into the expression for h to find the height of the box:
[tex]x=5 \implies h=\dfrac{75-(5)^2}{4(5)}=\dfrac{5}{2}=2.5[/tex]
Therefore, the dimensions that will produce a pen with maximum volume are:
- x = 5 ft
- y = 5 ft
- h = 2.5 ft
